Physicsgurus Q&A - Recent questions and answers in Velocity physics
https://physicsgurus.com/qa/velocity-physics
Powered by Question2AnswerAnswered: Are the equations for final velocity and final speed the same?
https://physicsgurus.com/88431/are-the-equations-for-final-velocity-and-final-speed-the-same?show=88432#a88432
<p>No, the equations for final velocity and final speed are not the same. While both concepts relate to the final state of motion, they differ in their definitions.</p><p>Final velocity refers to the velocity of an object at the end of a specific time interval or at a particular point in time. It is a vector quantity, which means it has both magnitude (speed) and direction. The equation for final velocity can be derived using the equation of motion:</p><p><span class="math math-inline"><span class="katex"><span class="katex-mathml">vf=vi+atv_f = v_i + at</span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height: 0.7167em; vertical-align: -0.2861em;"></span><span class="mord"><span class="mord mathnormal" style="margin-right: 0.03588em;">v</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height: 0.3361em;"><span style="top: -2.55em; margin-left: -0.0359em; margin-right: 0.05em;"><span class="pstrut" style="height: 2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mathnormal mtight" style="margin-right: 0.10764em;">f</span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height: 0.2861em;"><span></span></span></span></span></span></span><span class="mspace" style="margin-right: 0.2778em;"></span><span class="mrel">=</span><span class="mspace" style="margin-right: 0.2778em;"></span></span><span class="base"><span class="strut" style="height: 0.7333em; vertical-align: -0.15em;"></span><span class="mord"><span class="mord mathnormal" style="margin-right: 0.03588em;">v</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height: 0.3117em;"><span style="top: -2.55em; margin-left: -0.0359em; margin-right: 0.05em;"><span class="pstrut" style="height: 2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mathnormal mtight">i</span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height: 0.15em;"><span></span></span></span></span></span></span><span class="mspace" style="margin-right: 0.2222em;"></span><span class="mbin">+</span><span class="mspace" style="margin-right: 0.2222em;"></span></span><span class="base"><span class="strut" style="height: 0.6151em;"></span><span class="mord mathnormal">a</span><span class="mord mathnormal">t</span></span></span></span></span></p><p>Where:</p><ul><li><span class="math math-inline"><span class="katex"><span class="katex-mathml">vfv_f</span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height: 0.7167em; vertical-align: -0.2861em;"></span><span class="mord"><span class="mord mathnormal" style="margin-right: 0.03588em;">v</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height: 0.3361em;"><span style="top: -2.55em; margin-left: -0.0359em; margin-right: 0.05em;"><span class="pstrut" style="height: 2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mathnormal mtight" style="margin-right: 0.10764em;">f</span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height: 0.2861em;"><span></span></span></span></span></span></span></span></span></span></span> represents the final velocity</li><li><span class="math math-inline"><span class="katex"><span class="katex-mathml">viv_i</span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height: 0.5806em; vertical-align: -0.15em;"></span><span class="mord"><span class="mord mathnormal" style="margin-right: 0.03588em;">v</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height: 0.3117em;"><span style="top: -2.55em; margin-left: -0.0359em; margin-right: 0.05em;"><span class="pstrut" style="height: 2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mathnormal mtight">i</span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height: 0.15em;"><span></span></span></span></span></span></span></span></span></span></span> represents the initial velocity</li><li><span class="math math-inline"><span class="katex"><span class="katex-mathml">aa</span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height: 0.4306em;"></span><span class="mord mathnormal">a</span></span></span></span></span> represents the acceleration of the object</li><li><span class="math math-inline"><span class="katex"><span class="katex-mathml">tt</span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height: 0.6151em;"></span><span class="mord mathnormal">t</span></span></span></span></span> represents the time interval</li></ul><p>On the other hand, final speed refers to the magnitude of the final velocity, disregarding the direction. It is a scalar quantity, representing only the numerical value. The equation for final speed is derived by ignoring the direction component of the final velocity:</p><p><span class="math math-inline"><span class="katex"><span class="katex-mathml">vfinal speed=∣vf∣v_{ ext{final speed}} = |v_f|</span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height: 0.7167em; vertical-align: -0.2861em;"></span><span class="mord"><span class="mord mathnormal" style="margin-right: 0.03588em;">v</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height: 0.3361em;"><span style="top: -2.55em; margin-left: -0.0359em; margin-right: 0.05em;"><span class="pstrut" style="height: 2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight"><span class="mord text mtight"><span class="mord mtight">final speed</span></span></span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height: 0.2861em;"><span></span></span></span></span></span></span><span class="mspace" style="margin-right: 0.2778em;"></span><span class="mrel">=</span><span class="mspace" style="margin-right: 0.2778em;"></span></span><span class="base"><span class="strut" style="height: 1.0361em; vertical-align: -0.2861em;"></span><span class="mord">∣</span><span class="mord"><span class="mord mathnormal" style="margin-right: 0.03588em;">v</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height: 0.3361em;"><span style="top: -2.55em; margin-left: -0.0359em; margin-right: 0.05em;"><span class="pstrut" style="height: 2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mathnormal mtight" style="margin-right: 0.10764em;">f</span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height: 0.2861em;"><span></span></span></span></span></span></span><span class="mord">∣</span></span></span></span></span></p><p>In this equation, <span class="math math-inline"><span class="katex"><span class="katex-mathml">∣vf∣|v_f|</span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height: 1.0361em; vertical-align: -0.2861em;"></span><span class="mord">∣</span><span class="mord"><span class="mord mathnormal" style="margin-right: 0.03588em;">v</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height: 0.3361em;"><span style="top: -2.55em; margin-left: -0.0359em; margin-right: 0.05em;"><span class="pstrut" style="height: 2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mathnormal mtight" style="margin-right: 0.10764em;">f</span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height: 0.2861em;"><span></span></span></span></span></span></span><span class="mord">∣</span></span></span></span></span> denotes the magnitude (absolute value) of the final velocity vector.</p><p>Therefore, the equations for final velocity and final speed differ in terms of their vector nature. The final velocity accounts for both magnitude and direction, while the final speed only considers the magnitude.</p>Velocity physicshttps://physicsgurus.com/88431/are-the-equations-for-final-velocity-and-final-speed-the-same?show=88432#a88432Fri, 26 May 2023 10:03:25 +0000Answered: An aircraft takes off at 30 degrees with 500 km/hr. What will the vertical component of its velocity be?
https://physicsgurus.com/88321/aircraft-takes-degrees-with-what-vertical-component-velocity?show=88322#a88322
<p>To determine the vertical component of the velocity of the aircraft during takeoff, we need to use trigonometry and break down the given velocity into its horizontal and vertical components.</p><p>Given:</p><ul><li>Takeoff angle: 30 degrees</li><li>Takeoff velocity: 500 km/hr</li></ul><p>To find the vertical component of the velocity, we can use the following trigonometric relationship:</p><p>Vertical component = Velocity * sin(angle)</p><p>First, let's convert the velocity from km/hr to m/s to ensure consistent units:</p><p>500 km/hr * (1000 m/1 km) * (1 hr/3600 s) ≈ 138.89 m/s</p><p>Now we can calculate the vertical component:</p><p>Vertical component = 138.89 m/s * sin(30 degrees)</p><p>Using a calculator:</p><p>Vertical component ≈ 138.89 m/s * 0.5 ≈ 69.44 m/s</p><p>Therefore, the vertical component of the aircraft's velocity during takeoff will be approximately 69.44 m/s.</p>Velocity physicshttps://physicsgurus.com/88321/aircraft-takes-degrees-with-what-vertical-component-velocity?show=88322#a88322Fri, 26 May 2023 10:03:25 +0000Answered: Why two objects with the same mass that are launched with the same force don’t have the same velocity?
https://physicsgurus.com/88159/objects-with-same-mass-that-launched-with-same-force-velocity?show=88160#a88160
<p>Two objects with the same mass that are launched with the same force may not have the same velocity due to other factors that influence their motion. While the force applied to both objects is the same, several factors can affect their velocities:</p><ol><li><p>Initial conditions: Even with the same force, if the objects have different initial velocities, their resulting velocities will differ. For example, if one object is already moving before the force is applied, its final velocity will be different from the initially stationary object.</p></li><li><p>Air resistance: Objects moving through the air experience air resistance, which can vary depending on their shape, surface area, and velocity. If the two objects have different shapes or experience different levels of air resistance, their velocities will be affected differently.</p></li><li><p>Friction: If the objects are moving on surfaces with different frictional properties, such as different types of surfaces or different levels of roughness, the frictional forces acting on them will differ. This difference in frictional forces can cause variations in their velocities.</p></li><li><p>External influences: Other external factors, such as the presence of other forces (e.g., gravitational forces, magnetic forces) or interactions with other objects in the environment, can affect the velocities of the objects differently.</p></li></ol><p>In summary, while the force applied to two objects may be the same, their velocities can differ due to variations in their initial conditions, air resistance, friction, and other external influences acting upon them.</p>Velocity physicshttps://physicsgurus.com/88159/objects-with-same-mass-that-launched-with-same-force-velocity?show=88160#a88160Fri, 26 May 2023 10:03:25 +0000Answered: How do I find time with only final and initial velocity and acceleration?
https://physicsgurus.com/87827/how-find-time-with-only-final-initial-velocity-acceleration?show=87828#a87828
<p>To find the time it takes for an object to accelerate from an initial velocity to a final velocity when acceleration is constant, you can use the following equation of motion:</p><p>v = u + at</p><p>where: v = final velocity u = initial velocity a = acceleration t = time</p><p>Rearranging the equation, you can solve for time (t):</p><p>t = (v - u) / a</p><p>Simply substitute the known values of the final velocity (v), initial velocity (u), and acceleration (a) into the equation to calculate the time (t).</p>Velocity physicshttps://physicsgurus.com/87827/how-find-time-with-only-final-initial-velocity-acceleration?show=87828#a87828Fri, 26 May 2023 10:03:25 +0000Answered: To shoot a ball of radius 0.3 m and of mass 50 kg from the Earth’s surface to a 100 km orbit, what is the initial velocity and acceleration at which the ball lifts off?
https://physicsgurus.com/88619/shoot-radius-earths-surface-initial-velocity-acceleration?show=88620#a88620
<p>To calculate the initial velocity and acceleration required to launch a ball of radius 0.3 m and mass 50 kg from the Earth's surface to a 100 km orbit, we need to consider the gravitational force and the concept of escape velocity.</p><ol><li>Calculate the required escape velocity: The escape velocity is the minimum velocity needed to escape the gravitational pull of the Earth. It can be calculated using the formula: v_escape = sqrt((2 * G * M) / R) where G is the gravitational constant (approximately 6.67430 × 10^-11 m^3 kg^-1 s^-2), M is the mass of the Earth (approximately 5.972 × 10^24 kg), and R is the distance from the center of the Earth to the launch point (Earth's radius plus the altitude of the orbit).</li></ol><p>R = radius of the Earth + altitude of the orbit = 6,371 km + 100 km = 6,471 km = 6,471,000 m</p><p>Substituting the values into the escape velocity formula: v_escape = sqrt((2 * 6.67430 × 10^-11 * 5.972 × 10^24) / 6,471,000) ≈ 11,186 m/s</p><ol start="2"><li><p>Determine the initial velocity required: The initial velocity needed to lift off the ball must be greater than or equal to the escape velocity. Therefore, the initial velocity should be at least 11,186 m/s.</p></li><li><p>Calculate the acceleration at liftoff: At liftoff, the ball experiences two main forces: its weight (mg) acting downward and the upward thrust force (T). The net force acting on the ball is the difference between these forces: F_net = T - mg</p></li></ol><p>To overcome the ball's weight, the upward thrust force must be equal to or greater than the weight. Therefore, the acceleration at liftoff (a_liftoff) can be calculated using Newton's second law: F_net = ma_liftoff</p><p>Setting the thrust force equal to the weight: T = mg</p><p>Substituting this into the net force equation: F_net = mg - mg = 0</p><p>Therefore, at liftoff, the net force is zero, resulting in zero acceleration. The ball doesn't experience any acceleration once it leaves the ground, assuming no additional forces are acting on it.</p><p>In summary:</p><ul><li>The initial velocity required to launch the ball to a 100 km orbit is at least 11,186 m/s.</li><li>The acceleration at liftoff is 0 m/s².</li></ul>Velocity physicshttps://physicsgurus.com/88619/shoot-radius-earths-surface-initial-velocity-acceleration?show=88620#a88620Thu, 25 May 2023 10:03:25 +0000Answered: If an object is stationary and another object is approaching it with constant velocity, what will happen to the state of motion of the first object?
https://physicsgurus.com/88151/object-stationary-another-approaching-constant-velocity?show=88152#a88152
<p>If an object is stationary and another object is approaching it with a constant velocity, the state of motion of the first object will depend on various factors such as the nature of the interaction between the objects and the forces acting upon them.</p><p>In the absence of any external forces or interactions, the first object will remain stationary as the second object approaches it. This is based on Newton's first law of motion, which states that an object at rest tends to stay at rest unless acted upon by an external force.</p><p>However, if there is an interaction or force between the two objects, the first object may experience a change in its state of motion. For example:</p><ol><li><p>If the approaching object collides with the stationary object and imparts a force, the first object may start moving in the direction of the applied force or be displaced depending on the magnitude and direction of the force.</p></li><li><p>If there is a gravitational force between the objects, such as one object being attracted to another, the stationary object may be pulled towards the approaching object. The resulting motion will depend on the masses and distances involved.</p></li><li><p>If there are other forces acting on the objects, such as friction, air resistance, or electromagnetic forces, the stationary object may experience an external force that can cause it to move or change its state of motion.</p></li></ol><p>In summary, the state of motion of the first object, when approached by another object with constant velocity, will depend on the forces and interactions involved. In the absence of any external forces, the first object will remain stationary. However, the presence of forces or interactions can result in a change in the first object's state of motion.</p>Velocity physicshttps://physicsgurus.com/88151/object-stationary-another-approaching-constant-velocity?show=88152#a88152Thu, 25 May 2023 10:03:25 +0000Answered: When you ride a bicycle, in what direction is the angular velocity of the wheels?
https://physicsgurus.com/87483/when-you-ride-bicycle-what-direction-angular-velocity-wheels?show=87484#a87484
<p>When you ride a bicycle, the angular velocity of the wheels is in the same direction as the rotation of the wheels, which is clockwise for the front wheel and counterclockwise for the rear wheel (assuming you are seated on the bicycle and moving forward).</p><p>To clarify, when viewed from the side of the bicycle, the front wheel rotates clockwise, and the rear wheel rotates counterclockwise. The angular velocity vector points along the axis of rotation of the wheels, which in this case, is parallel to the axle of each wheel. Therefore, the angular velocity of both wheels points in the same direction as the rotation of the wheels.</p>Velocity physicshttps://physicsgurus.com/87483/when-you-ride-bicycle-what-direction-angular-velocity-wheels?show=87484#a87484Thu, 25 May 2023 10:03:25 +0000Answered: How will doubling the mass while keeping the initial velocity constant affect acceleration?
https://physicsgurus.com/88349/doubling-keeping-initial-velocity-constant-acceleration?show=88350#a88350
<p>According to Newton's second law of motion, the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Mathematically, this can be expressed as:</p><p>F = m * a</p><p>Where F represents the net force, m represents the mass, and a represents the acceleration.</p><p>If we keep the initial velocity constant and double the mass of an object, the acceleration will decrease. This can be understood by rearranging the equation for Newton's second law:</p><p>a = F / m</p><p>When the mass (m) is doubled while the net force (F) remains the same, the denominator in the equation increases, resulting in a smaller value for acceleration (a). Therefore, doubling the mass while keeping the initial velocity constant will cause a decrease in acceleration.</p>Velocity physicshttps://physicsgurus.com/88349/doubling-keeping-initial-velocity-constant-acceleration?show=88350#a88350Tue, 23 May 2023 10:03:25 +0000Answered: An object accelerates at 10 m/s², with an initial velocity of 5 m/s. What is the final velocity of the object, if the object took 5 seconds to accelerate?
https://physicsgurus.com/88045/accelerates-initial-velocity-velocity-seconds-accelerate?show=88046#a88046
<p>To determine the final velocity of an object undergoing constant acceleration, you can use the following equation:</p><p>v = u + at</p><p>Where: v is the final velocity u is the initial velocity a is the acceleration t is the time</p><p>Plugging in the values given in your question:</p><p>u = 5 m/s (initial velocity) a = 10 m/s² (acceleration) t = 5 s (time)</p><p>Using the equation:</p><p>v = u + at v = 5 m/s + 10 m/s² * 5 s v = 5 m/s + 50 m/s v = 55 m/s</p><p>Therefore, the final velocity of the object after 5 seconds of acceleration is 55 m/s.</p>Velocity physicshttps://physicsgurus.com/88045/accelerates-initial-velocity-velocity-seconds-accelerate?show=88046#a88046Tue, 23 May 2023 10:03:25 +0000Answered: A body starts from rest and uniformly accelerated for 10 seconds, so that its velocity becomes 100m/s. It then maintains this speed for 10 more seconds. What is the average velocity for this 20 seconds?
https://physicsgurus.com/87875/uniformly-accelerated-seconds-velocity-maintains-velocity?show=87876#a87876
<p>To find the average velocity for the 20 seconds, we can divide the total displacement by the total time taken.</p><p>In the first 10 seconds, the body starts from rest and uniformly accelerates until its velocity becomes 100 m/s. We can use the equation of motion:</p><p>v = u + at</p><p>Where: v = final velocity u = initial velocity (0 m/s, as it starts from rest) a = acceleration t = time</p><p>Rearranging the equation, we have:</p><p>a = (v - u) / t</p><p>Substituting the given values, we have:</p><p>a = (100 m/s - 0 m/s) / 10 s a = 10 m/s²</p><p>Since the body starts from rest, its initial velocity (u) is 0 m/s. We can calculate the displacement (s) using the equation:</p><p>s = ut + (1/2)at²</p><p>Substituting the values:</p><p>s = 0 + (1/2)(10 m/s²)(10 s)² s = 0 + 500 m s = 500 m</p><p>Now, in the next 10 seconds, the body maintains a constant speed of 100 m/s. The displacement during this time will be:</p><p>s = v * t s = 100 m/s * 10 s s = 1000 m</p><p>The total displacement for the 20 seconds is the sum of the displacements during the two intervals:</p><p>Total displacement = 500 m + 1000 m Total displacement = 1500 m</p><p>The average velocity is then calculated as the total displacement divided by the total time:</p><p>Average velocity = Total displacement / Total time Average velocity = 1500 m / 20 s Average velocity = 75 m/s</p><p>Therefore, the average velocity for the 20 seconds is 75 m/s.</p>Velocity physicshttps://physicsgurus.com/87875/uniformly-accelerated-seconds-velocity-maintains-velocity?show=87876#a87876Tue, 23 May 2023 10:03:25 +0000Answered: Two parallel rail tracks run north-south. Train A with a speed of 54k/hand B with speed of 90kmh. What is the relative velocity in ms of B with respect to A?
https://physicsgurus.com/87771/parallel-tracks-north-south-train-relative-velocity-respect?show=87772#a87772
<p>To find the relative velocity of Train B with respect to Train A, we need to consider their velocities as vectors and subtract them.</p><p>Given: Speed of Train A (vA) = 54 km/h Speed of Train B (vB) = 90 km/h</p><p>First, we convert the speeds from km/h to m/s: vA = 54 km/h * (1000 m/1 km) * (1/3600 h/1 s) ≈ 15 m/s vB = 90 km/h * (1000 m/1 km) * (1/3600 h/1 s) ≈ 25 m/s</p><p>Now, to find the relative velocity of B with respect to A, we subtract the velocity of A from the velocity of B: vRelative = vB - vA vRelative = 25 m/s - 15 m/s vRelative = 10 m/s</p><p>Therefore, the relative velocity of Train B with respect to Train A is 10 m/s.</p>Velocity physicshttps://physicsgurus.com/87771/parallel-tracks-north-south-train-relative-velocity-respect?show=87772#a87772Tue, 23 May 2023 10:03:25 +0000Answered: What are some examples of force, its direction, velocity, and acceleration?
https://physicsgurus.com/87541/what-some-examples-force-direction-velocity-acceleration?show=87542#a87542
<p>Sure! Here are some examples of forces along with their direction, velocity, and acceleration:</p><ol><li><p>Pushing a car:</p><ul><li>Force: The force exerted by a person on a car to move it.</li><li>Direction: The force is applied in the direction of the intended motion.</li><li>Velocity: The car's velocity increases as a result of the applied force.</li><li>Acceleration: The car accelerates in the direction of the force applied.</li></ul></li><li><p>Gravity:</p><ul><li>Force: The force of attraction between two objects due to their masses.</li><li>Direction: Gravity acts in the direction towards the center of mass of the objects.</li><li>Velocity: Gravity affects the velocity by causing objects to fall towards the ground.</li><li>Acceleration: Objects near the surface of the Earth experience an acceleration of approximately 9.8 m/s² downwards due to gravity.</li></ul></li><li><p>Friction:</p><ul><li>Force: The force that opposes the relative motion between two surfaces in contact.</li><li>Direction: The direction of friction force is opposite to the direction of motion or intended motion.</li><li>Velocity: Friction can decrease the velocity of an object in motion.</li><li>Acceleration: Friction can also cause a deceleration or negative acceleration.</li></ul></li><li><p>Magnetic force:</p><ul><li>Force: The force exerted on a charged particle or a current-carrying wire in a magnetic field.</li><li>Direction: The direction of the force depends on the charge, velocity, and magnetic field.</li><li>Velocity: The magnetic force can affect the particle's or wire's velocity, causing it to move in a curved path.</li><li>Acceleration: The magnetic force can result in an acceleration perpendicular to the initial velocity, altering the direction of motion.</li></ul></li><li><p>Tension in a string:</p><ul><li>Force: The force transmitted through a taut string or rope.</li><li>Direction: The direction of tension is along the string, away from the object applying the force.</li><li>Velocity: Tension does not directly affect velocity unless it is unbalanced by other forces.</li><li>Acceleration: If the tension is unbalanced, it can cause acceleration in the direction of the net force.</li></ul></li></ol><p>These examples illustrate different forces and their influence on the direction, velocity, and acceleration of objects in various scenarios.</p>Velocity physicshttps://physicsgurus.com/87541/what-some-examples-force-direction-velocity-acceleration?show=87542#a87542Tue, 23 May 2023 10:03:25 +0000Answered: When using the Bernoulli equation if the velocity term is incrementally increased while the constant is assumed to be unchanging, what happens to the pressure term?
https://physicsgurus.com/87231/bernoulli-equation-velocity-incrementally-increased-unchanging?show=87232#a87232
<p>According to the Bernoulli equation, which describes the conservation of energy in fluid flow, there is an inverse relationship between the velocity and pressure terms. If the velocity term is incrementally increased while assuming the constant term remains unchanged, the pressure term will decrease.</p><p>The Bernoulli equation can be expressed as:</p><p>P + (1/2)ρv^2 + ρgh = constant</p><p>Where: P is the pressure of the fluid ρ is the density of the fluid v is the velocity of the fluid g is the acceleration due to gravity h is the height of the fluid above a reference point</p><p>If we focus on the first two terms of the equation, P + (1/2)ρv^2, we can see that they represent the pressure and kinetic energy terms, respectively.</p><p>When the velocity term (v) is incrementally increased, assuming the constant term remains the same, the kinetic energy term (1/2)ρv^2 will increase. In order to maintain the constant value, the pressure term (P) must decrease. This decrease in pressure corresponds to a decrease in the static pressure of the fluid.</p><p>This relationship is often observed in fluid flow situations, such as when fluid flows through a constricted area, like a narrow pipe or nozzle. As the fluid's velocity increases in the constricted area, the pressure decreases according to the Bernoulli equation. This principle is utilized in applications like Venturi tubes, atomizers, and aircraft wings.</p>Velocity physicshttps://physicsgurus.com/87231/bernoulli-equation-velocity-incrementally-increased-unchanging?show=87232#a87232Tue, 23 May 2023 10:03:25 +0000Answered: The driver applies brakes on a bus moving at a speed of 40m/s and the velocity of the bus changes to 10m/ s in 6 seconds. What is the retardation in this case?
https://physicsgurus.com/87133/driver-applies-brakes-velocity-changes-seconds-retardation?show=87134#a87134
<p>To find the retardation, we need to calculate the change in velocity (Δv) and divide it by the time taken (t).</p><p>Given: Initial velocity (u) = 40 m/s Final velocity (v) = 10 m/s Time taken (t) = 6 seconds</p><p>The change in velocity (Δv) can be calculated as: Δv = v - u</p><p>Δv = 10 m/s - 40 m/s Δv = -30 m/s</p><p>The negative sign indicates that the velocity has decreased, or in other words, the bus is decelerating.</p><p>Now, we can calculate the retardation (a) using the formula: a = Δv / t</p><p>a = (-30 m/s) / (6 s) a = -5 m/s²</p><p>Therefore, the retardation in this case is -5 m/s². The negative sign indicates that it is a deceleration.</p>Velocity physicshttps://physicsgurus.com/87133/driver-applies-brakes-velocity-changes-seconds-retardation?show=87134#a87134Tue, 23 May 2023 10:03:25 +0000Answered: How can you show that the equation for impulse: Ft=mv_mu is dimensionally correct, where F=ma promptly U= initial velocity, v=final velocity, t=time?
https://physicsgurus.com/88419/equation-impulse-dimensionally-promptly-velocity-velocity?show=88420#a88420
<p>To show that the equation for impulse, Ft = m(v - u), is dimensionally correct, we need to analyze the dimensions of each term in the equation.</p><p>Let's break down the dimensions of each term:</p><p>F: Force (dimension: [M][L][T]^-2) t: Time (dimension: [T]) m: Mass (dimension: [M]) v: Final Velocity (dimension: [L][T]^-1) u: Initial Velocity (dimension: [L][T]^-1)</p><p>Now, let's substitute the dimensions into the equation and see if they are consistent:</p><p>Left-hand side (LHS): Ft = [M][L][T]^-2 [T] = [M][L][T]^-1</p><p>Right-hand side (RHS): m(v - u) = [M]([L][T]^-1 - [L][T]^-1) = [M][L][T]^-1</p><p>Both the LHS and RHS have the same dimensions of [M][L][T]^-1, which means they are consistent.</p><p>Therefore, the equation Ft = m(v - u) is dimensionally correct.</p>Velocity physicshttps://physicsgurus.com/88419/equation-impulse-dimensionally-promptly-velocity-velocity?show=88420#a88420Mon, 22 May 2023 10:03:25 +0000Answered: A wooden block of 12kg is hanging by a string on which a bullet of 100g has been shot. If the bullet sticks with the block and raised up to 5cm, what is the velocity of the bullet?
https://physicsgurus.com/88403/wooden-hanging-string-bullet-bullet-sticks-raised-velocity?show=88404#a88404
<p>To determine the velocity of the bullet, we can apply the principle of conservation of momentum.</p><p>The momentum before the collision is equal to the momentum after the collision. Assuming the bullet and the block are the only objects involved in the system:</p><p>Initial momentum = Final momentum</p><p>The initial momentum can be calculated as the product of the mass and velocity of the bullet before the collision, while the final momentum is the product of the combined mass (bullet + block) and the final velocity after the collision.</p><p>Given: Mass of the bullet (m₁) = 100 g = 0.1 kg Mass of the block (m₂) = 12 kg Initial velocity of the bullet (v₁) = ? Final velocity of the combined bullet and block (v₂) = 0 (since the system comes to rest)</p><p>Using the conservation of momentum equation:</p><p>m₁ * v₁ = (m₁ + m₂) * v₂</p><p>0.1 kg * v₁ = (0.1 kg + 12 kg) * 0</p><p>0.1 kg * v₁ = 0 kg</p><p>Since the mass of the bullet is nonzero and the final velocity is zero, it implies that the initial velocity of the bullet (v₁) must also be zero.</p><p>Hence, the velocity of the bullet before the collision is 0 m/s.</p>Velocity physicshttps://physicsgurus.com/88403/wooden-hanging-string-bullet-bullet-sticks-raised-velocity?show=88404#a88404Mon, 22 May 2023 10:03:25 +0000Answered: How do I find the maximum velocity if I’m given a maximum acceleration of 1.34 and a displacement of 806?
https://physicsgurus.com/88183/maximum-velocity-given-maximum-acceleration-displacement?show=88184#a88184
<p>To find the maximum velocity when given the maximum acceleration (a_max) and displacement (s), you can use the following equation:</p><p>v_max = sqrt(2 * a_max * s)</p><p>In this equation:</p><ul><li>v_max represents the maximum velocity</li><li>a_max represents the maximum acceleration</li><li>s represents the displacement</li></ul><p>Plugging in the given values, we have:</p><p>v_max = sqrt(2 * 1.34 * 806)</p><p>Calculating the expression inside the square root:</p><p>v_max = sqrt(2 * 1.34 * 806) v_max = sqrt(2165.12) v_max ≈ 46.54 m/s</p><p>Therefore, the maximum velocity, based on the given maximum acceleration of 1.34 and a displacement of 806, is approximately 46.54 m/s.</p>Velocity physicshttps://physicsgurus.com/88183/maximum-velocity-given-maximum-acceleration-displacement?show=88184#a88184Mon, 22 May 2023 10:03:25 +0000Answered: What is the shape of a velocity time graph for constant acceleration?
https://physicsgurus.com/87677/what-the-shape-velocity-time-graph-for-constant-acceleration?show=87678#a87678
<p>The shape of a velocity-time graph for constant acceleration is a straight line.</p><p>When an object undergoes constant acceleration, it means that its velocity changes by the same amount in equal time intervals. In other words, the object's acceleration remains constant throughout its motion.</p><p>On a velocity-time graph, the velocity is plotted on the y-axis, and time is plotted on the x-axis. Since acceleration is the rate of change of velocity, a constant acceleration corresponds to a linear increase or decrease in velocity over time.</p><p>If the object is undergoing constant positive acceleration (e.g., a car accelerating forward), the velocity-time graph will show a straight line with a positive slope. The slope represents the rate of change of velocity, which is the constant acceleration.</p><p>If the object is undergoing constant negative acceleration (e.g., a car decelerating or moving in the opposite direction), the velocity-time graph will show a straight line with a negative slope. Again, the slope represents the rate of change of velocity, which is the constant negative acceleration.</p><p>In both cases, the resulting velocity-time graph will be a straight line, indicating that the velocity changes uniformly over time due to the constant acceleration.</p>Velocity physicshttps://physicsgurus.com/87677/what-the-shape-velocity-time-graph-for-constant-acceleration?show=87678#a87678Mon, 22 May 2023 10:03:25 +0000Answered: A ball thrown straight upward returns to its original level in 2.74 s. A second ball is thrown at an angle of 30.0° above the horizontal. What is the initial speed of the second ball if it also returns to its original level in 2.75?
https://physicsgurus.com/87293/straight-returns-original-horizontal-initial-returns-original?show=87294#a87294
<p>To solve this problem, we can use the equations of motion for projectile motion. Let's consider the second ball thrown at an angle of 30.0° above the horizontal.</p><p>We'll assume that there is no air resistance and that the acceleration due to gravity is approximately 9.8 m/s² downward.</p><p>The given information is as follows:</p><ul><li>Time of flight (total time taken for the ball to return to its original level): t = 2.75 s</li><li>Angle of projection: θ = 30.0°</li></ul><p>To find the initial speed (magnitude) of the second ball, we need to break down its initial velocity into its horizontal and vertical components. We can use the equations:</p><p>Horizontal component: Vx = V₀ * cos(θ) Vertical component: Vy = V₀ * sin(θ)</p><p>Where V₀ is the initial speed.</p><p>Since the ball returns to its original level, we know that the vertical displacement is zero. The formula for the vertical displacement is given by:</p><p>Δy = Vy * t + (1/2) * g * t²</p><p>Since Δy is zero, we can simplify the equation to:</p><p>0 = Vy * t - (1/2) * g * t²</p><p>Now, substituting the expressions for Vy and Vx into the equation, we get:</p><p>0 = (V₀ * sin(θ)) * t - (1/2) * g * t²</p><p>Simplifying further:</p><p>0 = (V₀ * sin(θ)) * t - (4.9) * t²</p><p>Now, we can solve this quadratic equation for V₀. Rearranging and factoring out t:</p><p>0 = t * [(V₀ * sin(θ)) - (4.9) * t]</p><p>Since we know t ≠ 0, we can solve the expression in the brackets:</p><p>(V₀ * sin(θ)) - (4.9) * t = 0</p><p>V₀ * sin(θ) = (4.9) * t</p><p>Finally, solving for V₀:</p><p>V₀ = (4.9 * t) / sin(θ)</p><p>Plugging in the given values:</p><p>V₀ = (4.9 * 2.75) / sin(30.0°)</p><p>V₀ ≈ 26.95 m/s</p><p>Therefore, the initial speed of the second ball is approximately 26.95 m/s.</p>Velocity physicshttps://physicsgurus.com/87293/straight-returns-original-horizontal-initial-returns-original?show=87294#a87294Mon, 22 May 2023 10:03:25 +0000Answered: What is the final velocity of a body if it travels a given distance with uniform acceleration from the starting position?
https://physicsgurus.com/88753/velocity-travels-distance-acceleration-starting-position?show=88754#a88754
<p>If a body travels a given distance with uniform acceleration from its starting position, the final velocity can be determined using the following equation:</p><p>v^2 = u^2 + 2as</p><p>Where:</p><ul><li>v is the final velocity of the body</li><li>u is the initial velocity (which is typically zero if the body starts from rest)</li><li>a is the uniform acceleration</li><li>s is the distance traveled</li></ul><p>If the body starts from rest (u = 0), the equation simplifies to:</p><p>v^2 = 2as</p><p>To find the final velocity (v), you would take the square root of both sides of the equation:</p><p>v = √(2as)</p><p>It's important to note that the equation assumes the acceleration remains constant throughout the motion. If the acceleration is not constant or if there are other external factors involved, this equation may not provide an accurate estimation of the final velocity.</p>Velocity physicshttps://physicsgurus.com/88753/velocity-travels-distance-acceleration-starting-position?show=88754#a88754Sun, 21 May 2023 10:03:25 +0000Answered: Two photons are approaching each other. What is their relative velocity?
https://physicsgurus.com/88715/photons-approaching-each-other-what-their-relative-velocity?show=88716#a88716
<p>The relative velocity between two photons approaching each other is always equal to the speed of light (<span class="math math-inline"><span class="katex"><span class="katex-mathml">cc</span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height: 0.4306em;"></span><span class="mord mathnormal">c</span></span></span></span></span>). According to the theory of relativity, the speed of light in a vacuum is constant and is the maximum speed at which any information or particles can travel.</p><p>This means that regardless of the direction or relative motion of the photons, the speed at which they approach each other will always be <span class="math math-inline"><span class="katex"><span class="katex-mathml">cc</span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height: 0.4306em;"></span><span class="mord mathnormal">c</span></span></span></span></span>. This holds true for any observer, regardless of their own velocity or frame of reference.</p>Velocity physicshttps://physicsgurus.com/88715/photons-approaching-each-other-what-their-relative-velocity?show=88716#a88716Sun, 21 May 2023 10:03:25 +0000Answered: A stone is thrown vertically upward with the velocity of 25 m/s. How long does it take to reach the maximum height? What is the height? After what length of time will the stone hit the ground?
https://physicsgurus.com/88515/thrown-vertically-upward-velocity-maximum-height-height-length?show=88516#a88516
<p>To determine the time it takes for the stone to reach the maximum height, we can use the equation for vertical motion:</p><p>v = u + at</p><p>where: v = final velocity (0 m/s at the highest point, as the stone momentarily stops before falling back down) u = initial velocity (25 m/s) a = acceleration (acceleration due to gravity, -9.8 m/s², negative because it acts in the opposite direction of the upward motion) t = time taken to reach the maximum height</p><p>At the maximum height, the final velocity is zero, so we can rewrite the equation as:</p><p>0 = 25 m/s - 9.8 m/s² * t_max</p><p>Solving for t_max:</p><p>9.8 m/s² * t_max = 25 m/s</p><p>t_max = 25 m/s / 9.8 m/s² t_max ≈ 2.55 seconds</p><p>Therefore, it takes approximately 2.55 seconds for the stone to reach the maximum height.</p><p>To find the maximum height (h), we can use the kinematic equation:</p><p>v^2 = u^2 + 2as</p><p>where: v = final velocity (0 m/s) u = initial velocity (25 m/s) a = acceleration (acceleration due to gravity, -9.8 m/s²) s = displacement (height)</p><p>Rearranging the equation, we have:</p><p>0 = (25 m/s)^2 + 2 * (-9.8 m/s²) * h</p><p>625 m²/s² = 19.6 m/s² * h</p><p>h = 625 m²/s² / (19.6 m/s²) h ≈ 31.88 meters</p><p>Therefore, the maximum height reached by the stone is approximately 31.88 meters.</p><p>To determine the time it takes for the stone to hit the ground, we can consider the time it takes to reach the same height as the launch point, but in the opposite direction. Since the initial velocity is 25 m/s upwards, the final velocity when the stone hits the ground will be -25 m/s (downwards).</p><p>Using the equation:</p><p>v = u + at</p><p>where: v = final velocity (-25 m/s) u = initial velocity (25 m/s) a = acceleration (acceleration due to gravity, -9.8 m/s²) t = time taken to hit the ground</p><p>-25 m/s = 25 m/s - 9.8 m/s² * t_ground</p><p>Solving for t_ground:</p><p>-9.8 m/s² * t_ground = -50 m/s</p><p>t_ground = -50 m/s / -9.8 m/s² t_ground ≈ 5.1 seconds</p><p>Therefore, it takes approximately 5.1 seconds for the stone to hit the ground.</p>Velocity physicshttps://physicsgurus.com/88515/thrown-vertically-upward-velocity-maximum-height-height-length?show=88516#a88516Sun, 21 May 2023 10:03:25 +0000Answered: With an initial velocity of 7200 m/s, does a car accelerate m/S² for 10 seconds?
https://physicsgurus.com/88259/with-initial-velocity-7200-does-car-accelerate-for-seconds?show=88260#a88260
<p>To determine if a car accelerates at a rate of m/s² for 10 seconds, I would need the actual value of the acceleration. The unit "m/S²" you provided seems to be a combination of meters and seconds, but it's not clear what it represents without a specific value.</p><p>Acceleration is defined as the rate of change of velocity, and it is measured in units of meters per second squared (m/s²). It indicates how quickly the velocity of an object is changing over time.</p><p>If you provide the acceleration value, I can help you calculate the final velocity of the car after 10 seconds using the given initial velocity and acceleration.</p>Velocity physicshttps://physicsgurus.com/88259/with-initial-velocity-7200-does-car-accelerate-for-seconds?show=88260#a88260Sun, 21 May 2023 10:03:25 +0000Answered: An object of mass 4kg is moving towards east at velocity 6m/s . It collides and sticks to a 6kg moving with a velocity of 5m/s in the same direction. How much kinetic energy was lost in a collision?
https://physicsgurus.com/88185/towards-velocity-collides-velocity-direction-kinetic-collision?show=88186#a88186
<p>To calculate the kinetic energy lost in the collision, we need to determine the initial and final kinetic energies of the system.</p><p>The initial kinetic energy of the system is the sum of the kinetic energies of the two objects before the collision. The kinetic energy (KE) of an object is given by the equation KE = (1/2)mv^2, where m is the mass of the object and v is its velocity.</p><p>For the first object with a mass of 4 kg and velocity of 6 m/s, the initial kinetic energy is: KE1 = (1/2) * 4 kg * (6 m/s)^2 = 72 J</p><p>For the second object with a mass of 6 kg and velocity of 5 m/s, the initial kinetic energy is: KE2 = (1/2) * 6 kg * (5 m/s)^2 = 75 J</p><p>The final kinetic energy of the system is the kinetic energy of the combined mass after the collision.</p><p>Since the two objects stick together after the collision, they become one object with a total mass of 4 kg + 6 kg = 10 kg.</p><p>The final velocity of the combined object can be found using the principle of conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision. Since the two objects are moving in the same direction, the momentum equation is:</p><p>(mass1 * velocity1) + (mass2 * velocity2) = (mass1 + mass2) * velocity_final</p><p>(4 kg * 6 m/s) + (6 kg * 5 m/s) = 10 kg * velocity_final</p><p>24 kg·m/s + 30 kg·m/s = 10 kg * velocity_final</p><p>54 kg·m/s = 10 kg * velocity_final</p><p>velocity_final = 54 kg·m/s / 10 kg velocity_final = 5.4 m/s</p><p>The final kinetic energy (KE_final) of the system is given by: KE_final = (1/2) * total_mass * velocity_final^2</p><p>KE_final = (1/2) * 10 kg * (5.4 m/s)^2 KE_final = 145.8 J</p><p>The kinetic energy lost in the collision is the difference between the initial kinetic energy and the final kinetic energy:</p><p>Kinetic energy lost = Initial kinetic energy - Final kinetic energy Kinetic energy lost = (KE1 + KE2) - KE_final Kinetic energy lost = (72 J + 75 J) - 145.8 J Kinetic energy lost = 147 J - 145.8 J Kinetic energy lost ≈ 1.2 J</p><p>Therefore, approximately 1.2 Joules of kinetic energy was lost in the collision.</p>Velocity physicshttps://physicsgurus.com/88185/towards-velocity-collides-velocity-direction-kinetic-collision?show=88186#a88186Sun, 21 May 2023 10:03:25 +0000Answered: How do you calculate the ratio of average to instantaneous velocity (or speed)?
https://physicsgurus.com/87829/how-calculate-ratio-average-instantaneous-velocity-speed?show=87830#a87830
<p>The ratio of average velocity to instantaneous velocity (or speed) can be calculated using the following formula:</p><p>Ratio = Average Velocity / Instantaneous Velocity</p><p>To find the average velocity, you divide the displacement (change in position) by the total time interval:</p><p>Average Velocity = Displacement / Time</p><p>To calculate the instantaneous velocity (or speed) at a specific point in time, you would need the derivative of the position function with respect to time. In other words, you take the derivative of the position function to obtain the velocity function and evaluate it at the desired time:</p><p>Instantaneous Velocity = d(position) / dt</p><p>Once you have the average velocity and instantaneous velocity (or speed), you can then calculate their ratio using the formula mentioned above.</p>Velocity physicshttps://physicsgurus.com/87829/how-calculate-ratio-average-instantaneous-velocity-speed?show=87830#a87830Sun, 21 May 2023 10:03:25 +0000Answered: If the velocity of a car changes from 18 kmph to 72 kmph in 30 seconds, then what is the acceleration in km/h square?
https://physicsgurus.com/87509/velocity-changes-from-kmph-kmph-seconds-acceleration-square?show=87510#a87510
<p>To calculate the acceleration, we need to convert the initial and final velocities to the same unit (km/h) and divide the change in velocity by the time taken.</p><p>Given: Initial velocity = 18 km/h Final velocity = 72 km/h Time = 30 seconds</p><p>Converting the velocities to km/h: Initial velocity = 18 km/h Final velocity = 72 km/h</p><p>Change in velocity = Final velocity - Initial velocity Change in velocity = 72 km/h - 18 km/h Change in velocity = 54 km/h</p><p>Acceleration = Change in velocity / Time Acceleration = 54 km/h / 30 seconds</p><p>Simplifying the units: Acceleration = (54 km/h) / (30/3600) h Acceleration = (54 km/h) / (0.008333 h) Acceleration = 6480 km/h² / h</p><p>Therefore, the acceleration of the car is 6480 km/h².</p>Velocity physicshttps://physicsgurus.com/87509/velocity-changes-from-kmph-kmph-seconds-acceleration-square?show=87510#a87510Sun, 21 May 2023 10:03:25 +0000A boat moves north across a river at 15m/s. There is a current moving east at 15m/s. What is the magnitude and direction of his resultant velocity?
https://physicsgurus.com/87357/across-current-moving-magnitude-direction-resultant-velocity
Velocity physicshttps://physicsgurus.com/87357/across-current-moving-magnitude-direction-resultant-velocitySat, 20 May 2023 10:03:25 +0000What is the best program for calculating stream discharge using cross-section and velocity data? I have multiple velocity data points at various depths and hoping to avoid doing these calculations by hand.
https://physicsgurus.com/87253/calculating-discharge-velocity-multiple-velocity-calculations
Velocity physicshttps://physicsgurus.com/87253/calculating-discharge-velocity-multiple-velocity-calculationsSat, 20 May 2023 10:03:25 +0000Answered: What is the velocity at which an object can start moving in space when it is at the "zero point"?
https://physicsgurus.com/88547/what-velocity-which-object-start-moving-space-when-zero-point?show=88548#a88548
<p>In the context of space, the velocity required for an object to start moving from a stationary position at the "zero point" depends on various factors.</p><ol><li><p>Gravitational force: If the object is within the gravitational field of a massive celestial body, such as a planet or a star, the required velocity would depend on the escape velocity. The escape velocity is the minimum velocity an object needs to overcome the gravitational pull and escape the gravitational field. It is given by the equation:</p><p>v = sqrt(2 * G * M / r),</p><p>where: v is the escape velocity, G is the gravitational constant, M is the mass of the celestial body, and r is the distance between the object and the center of the celestial body.</p><p>The specific escape velocity varies depending on the mass and radius of the celestial body.</p></li><li><p>Orbital mechanics: If the object is already in orbit around a celestial body, the velocity required to start moving away from the "zero point" would depend on the orbital velocity. The orbital velocity is the velocity required for an object to maintain a stable orbit around a celestial body without falling back to the surface. It is determined by the gravitational force between the object and the celestial body. The specific orbital velocity depends on the mass and radius of the celestial body.</p></li><li><p>Thrust and propulsion: In space, if an object has access to a propulsion system, such as rocket engines, the velocity required to start moving from the "zero point" would depend on the thrust provided by the propulsion system. The object needs to generate sufficient thrust to overcome any external forces acting on it, such as gravity or atmospheric drag, if present.</p></li></ol><p>In summary, the velocity required for an object to start moving in space from the "zero point" depends on factors such as the gravitational force, orbital mechanics, and the presence of a propulsion system. The specific velocity needed can vary depending on the specific circumstances and the properties of the celestial body involved.</p>Velocity physicshttps://physicsgurus.com/88547/what-velocity-which-object-start-moving-space-when-zero-point?show=88548#a88548Sat, 20 May 2023 10:03:25 +0000Answered: If a body is projected with a velocity of 36 km/hr in an upward direction, what is its momentum at maximum height?
https://physicsgurus.com/87685/projected-velocity-upward-direction-momentum-maximum-height?show=87686#a87686
<p>To find the momentum of a body at its maximum height, we need to first calculate its velocity at that point. Given that the body is projected with a velocity of 36 km/hr in an upward direction, we can determine its final velocity at the maximum height using the principles of projectile motion.</p><p>At the maximum height, the body momentarily comes to rest before changing its direction of motion. Therefore, the final velocity (v_f) at the maximum height is 0 km/hr.</p><p>To find the momentum (p) at the maximum height, we can use the formula:</p><p>p = m * v_f</p><p>where: p is the momentum, m is the mass of the body, v_f is the final velocity.</p><p>Since the problem does not provide information about the mass of the body, we cannot determine the exact value of the momentum without knowing the mass. Momentum is directly proportional to mass, so without the mass, we cannot calculate the momentum.</p><p>However, if you have the mass of the body, you can substitute it into the formula along with the final velocity of 0 km/hr to calculate the momentum at the maximum height.</p>Velocity physicshttps://physicsgurus.com/87685/projected-velocity-upward-direction-momentum-maximum-height?show=87686#a87686Sat, 20 May 2023 10:03:25 +0000Answered: If drift velocity is constant in the circuit that kinetic energy is constant, why are there drops or changes in potential difference across components?
https://physicsgurus.com/87467/velocity-constant-constant-potential-difference-components?show=87468#a87468
<p>In a circuit, the drift velocity of electrons is indeed constant in a given wire. However, the potential difference (voltage) across components can still change because different components in the circuit offer resistance to the flow of electrons.</p><p>When an electric current flows through a circuit, it encounters resistance, which causes a drop in voltage across the components. This drop in voltage is related to the energy loss in the circuit due to the resistance.</p><p>According to Ohm's law, the voltage across a component (V) is equal to the current (I) multiplied by the resistance (R): V = I * R. So, as the current flows through a component with resistance, the voltage across that component decreases.</p><p>This voltage drop occurs because some of the electrical energy is converted into other forms of energy, such as heat or light, depending on the component. For example, in a resistor, the energy is dissipated as heat due to the resistance of the material.</p><p>In circuits with multiple components, such as resistors, capacitors, or diodes, each component offers a different resistance to the flow of current, resulting in different voltage drops across them.</p><p>It's important to note that the total energy of the electrons, including their kinetic energy, remains constant throughout the circuit. However, the distribution of that energy can change as the electrons encounter resistance and voltage drops across the components.</p><p>In summary, even though the drift velocity of electrons remains constant, changes in potential difference (voltage) across components occur due to the resistance encountered by the current flow, resulting in voltage drops across those components.</p>Velocity physicshttps://physicsgurus.com/87467/velocity-constant-constant-potential-difference-components?show=87468#a87468Sat, 20 May 2023 10:03:25 +0000Does a bullet with 4x the powder have double or quadruple the velocity?
https://physicsgurus.com/87889/does-bullet-with-the-powder-have-double-quadruple-velocity
Velocity physicshttps://physicsgurus.com/87889/does-bullet-with-the-powder-have-double-quadruple-velocityFri, 19 May 2023 10:03:25 +0000Answered: A projectile thrown from a point in a horizontal plane comes back to the plane in 4 seconds at a distance of 60m in front of the point of projection. What is the velocity of projection?
https://physicsgurus.com/88617/projectile-horizontal-distance-projection-velocity-projection?show=88618#a88618
<p>To determine the velocity of projection, we can use the fact that the time of flight of a projectile is twice the time it takes for the projectile to reach its maximum height. We'll also assume that there is no air resistance.</p><p>Let's denote the velocity of projection as 'v' and the time it takes for the projectile to reach its maximum height as 't'. The time of flight is then 2t.</p><p>Since the projectile reaches a maximum height and returns to the same horizontal plane, the vertical component of velocity at the highest point is zero.</p><p>Using the equations of motion for the vertical direction:</p><p>At the highest point: v_y = 0 v_y = u_y + a_y * t 0 = v * sinθ - g * t t = v * sinθ / g</p><p>Where: v_y is the vertical component of velocity u_y is the initial vertical component of velocity (which is v * sinθ) a_y is the acceleration due to gravity (-g) t is the time it takes to reach maximum height</p><p>Now, considering the horizontal motion, we know that the horizontal displacement is 60 meters and the time of flight is 4 seconds:</p><p>x = u_x * t 60 = v * cosθ * (2t)</p><p>Since t = v * sinθ / g, we can substitute this expression into the equation:</p><p>60 = v * cosθ * (2 * (v * sinθ / g))</p><p>Simplifying the equation:</p><p>60 = (2v² * sinθ * cosθ) / g</p><p>Using the trigonometric identity sin(2θ) = 2sinθ * cosθ, we can rewrite the equation as:</p><p>60 = (v² * sin(2θ)) / g</p><p>Now we can solve for v:</p><p>v² = (60 * g) / sin(2θ) v = √((60 * g) / sin(2θ))</p><p>Finally, we need to find the angle θ. Since the vertical component of velocity at the highest point is zero, we know that the time taken to reach the highest point is half of the total time of flight. Thus:</p><p>t = 4 seconds / 2 = 2 seconds</p><p>Now we can substitute this time into the equation for t above:</p><p>2 = v * sinθ / g</p><p>Simplifying:</p><p>sinθ = (2 * g) / v θ = arcsin((2 * g) / v)</p><p>Substituting the expression for v from the previous equation, we get:</p><p>θ = arcsin((2 * g) / √((60 * g) / sin(2θ)))</p><p>Solving for θ numerically using an equation solver would yield the angle of projection.</p>Velocity physicshttps://physicsgurus.com/88617/projectile-horizontal-distance-projection-velocity-projection?show=88618#a88618Fri, 19 May 2023 10:03:25 +0000Answered: A ball is thrown horizontally 20 m above the ground with a velocity of 5 m/s. How far from the base of the cliff will the ball land?
https://physicsgurus.com/88483/ball-thrown-horizontally-above-ground-with-velocity-cliff?show=88484#a88484
<p>To solve this problem, we can use the equations of motion for projectile motion. Since the ball is thrown horizontally, the initial vertical velocity is 0 m/s. We can assume that the acceleration due to gravity is 9.8 m/s² downward.</p><p>We can use the equation: h = ut + (1/2)gt²</p><p>Where: h = vertical displacement (20 m) u = initial vertical velocity (0 m/s) g = acceleration due to gravity (-9.8 m/s²) t = time of flight</p><p>Since the ball is thrown horizontally, the time taken to reach the ground is the same as the time of flight. Rearranging the equation, we get:</p><p>t = sqrt(2h/g)</p><p>Substituting the values, we have: t = sqrt(2 * 20 / 9.8) ≈ 2.02 s</p><p>Now, we can use the formula for horizontal distance traveled: d = horizontal velocity × time of flight</p><p>Since the ball is thrown horizontally, the horizontal velocity is 5 m/s. Substituting the values, we have: d = 5 m/s × 2.02 s ≈ 10.1 m</p><p>Therefore, the ball will land approximately 10.1 meters from the base of the cliff.</p>Velocity physicshttps://physicsgurus.com/88483/ball-thrown-horizontally-above-ground-with-velocity-cliff?show=88484#a88484Fri, 19 May 2023 10:03:25 +0000Answered: A block starts from rest and slides down an inclined plane with an acceleration of 2 m/s^2. How far will it go during the first 2 seconds? What will be its velocity at the end of the 2nd second?
https://physicsgurus.com/88315/starts-slides-inclined-acceleration-during-seconds-velocity?show=88316#a88316
<p>To solve this problem, we can use the equations of motion for uniformly accelerated linear motion. The key equation we will need is:</p><p>s = ut + (1/2)at^2</p><p>where: s is the distance traveled, u is the initial velocity, t is the time, and a is the acceleration.</p><p>Given: Initial velocity (u) = 0 m/s (since the block starts from rest), Acceleration (a) = 2 m/s^2, and Time (t) = 2 seconds.</p><p>Let's calculate the distance traveled during the first 2 seconds:</p><p>s = ut + (1/2)at^2 s = 0 * 2 + (1/2) * 2 * (2^2) s = 0 + (1/2) * 2 * 4 s = 0 + 1 * 4 s = 4 meters</p><p>Therefore, the block will travel a distance of 4 meters during the first 2 seconds.</p><p>Now, let's find the velocity of the block at the end of the 2nd second. We can use the equation:</p><p>v = u + at</p><p>Given: Initial velocity (u) = 0 m/s (since the block starts from rest), Acceleration (a) = 2 m/s^2, and Time (t) = 2 seconds.</p><p>v = u + at v = 0 + 2 * 2 v = 0 + 4 v = 4 m/s</p><p>Therefore, the velocity of the block at the end of the 2nd second will be 4 m/s.</p>Velocity physicshttps://physicsgurus.com/88315/starts-slides-inclined-acceleration-during-seconds-velocity?show=88316#a88316Fri, 19 May 2023 10:03:25 +0000Answered: For the acceleration of a particle moving along a straight line is a=-0.2v*2m/s*2 where vis in m/s. If the initial velocity v=80M/s at t=0, what is the velocity when t=2 seconds?
https://physicsgurus.com/88101/acceleration-particle-straight-initial-velocity-velocity?show=88102#a88102
<p>To find the velocity of the particle when t = 2 seconds, we can use the given acceleration equation and the initial velocity.</p><p>Given: Acceleration, a = -0.2v^2 m/s^2 Initial velocity, v = 80 m/s Time, t = 2 seconds</p><p>We can solve this problem using calculus and integration, as the acceleration is a function of velocity. We'll integrate the acceleration equation with respect to time to obtain the velocity equation.</p><p>dv/dt = a dv = a dt ∫dv = ∫a dt</p><p>Integrating both sides of the equation, we get:</p><p>∫dv = ∫(-0.2v^2) dt</p><p>Integrating the left side gives us v, and integrating the right side gives us -0.2 * ∫v^2 dt.</p><p>v = -0.2 * (v^3/3) + C</p><p>To find the constant of integration, C, we can substitute the initial velocity and time:</p><p>v = -0.2 * (v^3/3) + C 80 = -0.2 * (80^3/3) + C 80 = -0.2 * (512000/3) + C 80 = -34133.333 + C C = 34213.333</p><p>Now we can substitute t = 2 seconds into the equation to find the velocity at that time:</p><p>v = -0.2 * (v^3/3) + 34213.333 v = -0.2 * (v^3/3) + 34213.333</p><p>This equation cannot be solved algebraically, so we can use numerical methods to approximate the velocity. One such method is the Newton-Raphson method. However, since the velocity equation is complex, I am unable to provide a specific numerical solution without using a numerical computing environment or calculator.</p>Velocity physicshttps://physicsgurus.com/88101/acceleration-particle-straight-initial-velocity-velocity?show=88102#a88102Fri, 19 May 2023 10:03:25 +0000Answered: How can the speed be zero, if velocity is zero?
https://physicsgurus.com/88059/how-can-the-speed-be-zero-if-velocity-is-zero?show=88060#a88060
<p>I apologize for any confusion caused by my previous response. You are correct in pointing out the discrepancy. If the velocity of an object is zero, it means the object is at rest and not in motion. In such a case, the speed of the object would also be zero since speed is defined as the magnitude of velocity, disregarding its direction.</p><p>To clarify, when an object has constant velocity, it means that it is moving in a straight line at a constant speed and in a specific direction. If the velocity is zero, the object is stationary, and there is no motion or change in position occurring. Therefore, the speed would also be zero in this case.</p><p>Once again, I apologize for the oversight and any confusion it may have caused. If you have any further questions, please feel free to ask.</p>Velocity physicshttps://physicsgurus.com/88059/how-can-the-speed-be-zero-if-velocity-is-zero?show=88060#a88060Fri, 19 May 2023 10:03:25 +0000Answered: What is the relationship between mass, orbit distance and orbital velocity of an object? If there is a relationship, how can it be calculated precisely?
https://physicsgurus.com/87503/relationship-distance-relationship-calculated-precisely?show=87504#a87504
<p>The relationship between mass, orbit distance, and orbital velocity of an object can be described by Kepler's third law of planetary motion. This law states that the square of the orbital period (T) of a planet or satellite is proportional to the cube of its average distance from the central object (r) around which it orbits. Mathematically, it can be expressed as:</p><p>T^2 ∝ r^3</p><p>The orbital velocity (V) of an object can be calculated using the formula:</p><p>V = (2πr) / T</p><p>where V represents the orbital velocity, r is the average distance from the central object, and T is the orbital period.</p><p>However, it's important to note that the above relationship assumes a two-body system where the mass of the orbiting object is negligible compared to the central object. In reality, when the mass of the orbiting object is significant (e.g., a planet orbiting a star), the calculation becomes more complex and involves the gravitational force between the objects. The precise calculation of orbital velocity in such cases requires considering the masses of both objects and employing the laws of gravitational dynamics, such as Newton's law of universal gravitation and the concept of centripetal force.</p><p>In summary, the relationship between mass, orbit distance, and orbital velocity is intricate and depends on the specific scenario, incorporating principles from Kepler's laws and gravitational dynamics.</p>Velocity physicshttps://physicsgurus.com/87503/relationship-distance-relationship-calculated-precisely?show=87504#a87504Fri, 19 May 2023 10:03:25 +0000Answered: What is the formula for finding the final height of a ball thrown vertically upwards with constant acceleration?
https://physicsgurus.com/87437/formula-finding-vertically-upwards-constant-acceleration?show=87438#a87438
<p>The formula for finding the final height (h) of a ball thrown vertically upwards with constant acceleration can be derived using the equations of motion.</p><p>When a ball is thrown vertically upwards, it experiences a constant acceleration due to gravity acting in the opposite direction to its motion. The acceleration is typically denoted as "g" and has a value of approximately 9.8 m/s² near the surface of the Earth.</p><p>The relevant equation for finding the final height is:</p><p>h = (v² - u²) / (2g)</p><p>Where: h = final height v = final velocity u = initial velocity g = acceleration due to gravity</p><p>When the ball reaches its highest point, the final velocity is 0 m/s because it momentarily comes to a stop before starting to descend. Therefore, the equation becomes:</p><p>h = (0² - u²) / (2g) h = -u² / (2g)</p><p>Note that the negative sign arises because the initial velocity (u) is directed upwards, while the convention for height is typically measured upwards from a reference point. The negative sign indicates that the final height is below the initial height.</p><p>So, the formula for finding the final height of a ball thrown vertically upwards with constant acceleration is:</p><p>h = -u² / (2g)</p>Velocity physicshttps://physicsgurus.com/87437/formula-finding-vertically-upwards-constant-acceleration?show=87438#a87438Fri, 19 May 2023 10:03:25 +0000Answered: On what thing velocity of electrons depends?
https://physicsgurus.com/87245/on-what-thing-velocity-of-electrons-depends?show=87246#a87246
<p>The velocity of electrons can depend on various factors. Here are a few factors that can influence the velocity of electrons:</p><ol><li><p>Electric Field: When electrons are subjected to an electric field, they experience a force and can accelerate or decelerate depending on the direction of the field. The magnitude and direction of the electric field can impact the velocity of electrons.</p></li><li><p>Potential Difference: The potential difference, also known as voltage, across a circuit or an electric field can affect the velocity of electrons. Higher voltage can result in higher electron velocities.</p></li><li><p>Temperature: In conductors or semiconductors, temperature can influence the velocity of electrons. As temperature increases, the electrons gain more thermal energy and can move with higher velocities due to increased kinetic energy.</p></li><li><p>Material Properties: The properties of the material through which the electrons are moving can also affect their velocity. For example, in conductors, electrons can move more freely and have higher velocities compared to semiconductors or insulators.</p></li><li><p>External Forces: External forces such as magnetic fields can influence the velocity of electrons, as they may experience a magnetic force that affects their motion.</p></li></ol><p>It's important to note that the velocity of individual electrons in a conductor or a circuit can vary, but the average velocity (known as drift velocity) is typically much lower due to frequent collisions with atoms or other electrons in the material.</p>Velocity physicshttps://physicsgurus.com/87245/on-what-thing-velocity-of-electrons-depends?show=87246#a87246Fri, 19 May 2023 10:03:25 +0000Answered: Can accelerated velocity change suddenly? And can velocity change suddenly?
https://physicsgurus.com/87151/accelerated-velocity-change-suddenly-velocity-change-suddenly?show=87152#a87152
<p>Acceleration is defined as the rate of change of velocity over time. In classical physics, acceleration refers to a continuous change in velocity. According to the laws of physics, velocity cannot change instantaneously or suddenly without an external force or an impulse acting on an object.</p><p>However, in certain situations, it may appear as if velocity or acceleration changes suddenly due to practical limitations in measurements or the presence of very short-duration events. For example, when observing the motion of an object with limited precision or using discrete measurements, the velocity or acceleration values may appear to change abruptly. But in reality, there would still be an underlying continuous process of change.</p><p>It's important to note that abrupt changes in velocity or acceleration often indicate the presence of external forces or events such as collisions, explosions, or instantaneous impulses. In such cases, the change in velocity or acceleration occurs rapidly but not instantaneously.</p>Velocity physicshttps://physicsgurus.com/87151/accelerated-velocity-change-suddenly-velocity-change-suddenly?show=87152#a87152Fri, 19 May 2023 10:03:25 +0000A cannonball is fired at an initial velocity of 85.76m/s elevated at an angle of 45° 1. What is the initial velocity of the cannon ball? 4. What is the maximum height reached by the cannonball? 5. What is the initial horizontal velocity?
https://physicsgurus.com/88433/cannonball-velocity-elevated-velocity-cannonball-horizontal
Velocity physicshttps://physicsgurus.com/88433/cannonball-velocity-elevated-velocity-cannonball-horizontalThu, 18 May 2023 10:03:25 +0000How long does it take for an automobile to travel ninety kilometers at an average velocity of thirty kilometers per hour (round to two decimal places)?
https://physicsgurus.com/88207/automobile-kilometers-average-velocity-kilometers-decimal
Velocity physicshttps://physicsgurus.com/88207/automobile-kilometers-average-velocity-kilometers-decimalThu, 18 May 2023 10:03:25 +0000Answered: What is the fastest possible velocity of a body that can be achieved in space?
https://physicsgurus.com/88745/what-the-fastest-possible-velocity-body-that-achieved-space?show=88746#a88746
<p>In the context of special relativity, the theory developed by Albert Einstein, the fastest possible velocity in space is the speed of light in a vacuum, denoted by the symbol "c." The speed of light is approximately 299,792,458 meters per second (or about 299,792 kilometers per second).</p><p>According to the theory of special relativity, as an object with mass accelerates, its relativistic mass increases, and the energy required to continue accelerating it also increases. As the object approaches the speed of light, its relativistic mass becomes infinite, and it would require an infinite amount of energy to accelerate it further.</p><p>Thus, based on our current understanding of physics, no object with mass can reach or exceed the speed of light. This concept is encapsulated in the principle known as the "cosmic speed limit" or "Einstein's cosmic speed limit," which states that information, energy, or matter cannot travel faster than the speed of light.</p><p>It's worth noting that special relativity applies to objects with mass. However, there are hypothetical particles called "tachyons" that are sometimes discussed in physics, which, if they exist, would always travel faster than the speed of light. However, tachyons are purely speculative and have not been observed or confirmed in any experimental context.</p>Velocity physicshttps://physicsgurus.com/88745/what-the-fastest-possible-velocity-body-that-achieved-space?show=88746#a88746Thu, 18 May 2023 10:03:25 +0000Answered: An arrow is shot from the top of a 30m cliff straight up in the air at an initial speed of 15m/s. What is the height at which it is moving with 8m/S?
https://physicsgurus.com/88649/arrow-shot-cliff-straight-initial-speed-height-which-moving?show=88650#a88650
<p>To determine the height at which the arrow is moving with a speed of 8 m/s, we can use the principles of projectile motion. Since the arrow is shot straight up, we only need to consider the vertical motion.</p><p>Let's denote:</p><ul><li>u as the initial velocity (15 m/s)</li><li>v as the final velocity (8 m/s)</li><li>g as the acceleration due to gravity (-9.8 m/s², taking into account its downward direction)</li><li>h as the height we want to find</li></ul><p>Using the equation for the final velocity in vertical motion: v² = u² + 2gh</p><p>Rearranging the equation to solve for h: h = (v² - u²) / (2g)</p><p>Plugging in the given values: h = (8² - 15²) / (2 * -9.8)</p><p>Calculating the expression: h = (64 - 225) / (-19.6) h = -161 / -19.6 h ≈ 8.22 meters</p><p>Therefore, when the arrow is moving with a speed of 8 m/s, it is at a height of approximately 8.22 meters above the initial launch point.</p>Velocity physicshttps://physicsgurus.com/88649/arrow-shot-cliff-straight-initial-speed-height-which-moving?show=88650#a88650Thu, 18 May 2023 10:03:25 +0000Answered: A cliff diver jumps from a ledge 96 feet above the ocean with an initial upward velocity of 16 feet per second. How long will it take until the diver enters the water?
https://physicsgurus.com/88209/cliff-diver-jumps-ledge-initial-upward-velocity-second-enters?show=88210#a88210
<p>To solve this problem, we can use the kinematic equation for vertical motion:</p><p>h = ut + (1/2)gt^2</p><p>where: h = height (distance) fallen u = initial velocity g = acceleration due to gravity (approximately -32 feet per second squared) t = time</p><p>In this case, the height fallen is 96 feet, the initial velocity is 16 feet per second, and the acceleration due to gravity is -32 feet per second squared. We need to find the time it takes until the diver enters the water, so we can set h = 0 and solve for t.</p><p>0 = ut + (1/2)gt^2</p><p>0 = 16t + (1/2)(-32)t^2</p><p>0 = 16t - 16t^2</p><p>Rearranging the equation:</p><p>16t^2 - 16t = 0</p><p>Factor out 16t:</p><p>16t(t - 1) = 0</p><p>This equation has two solutions: t = 0 and t = 1. However, t = 0 represents the initial time when the diver is still on the ledge, so we can ignore that solution.</p><p>Therefore, the diver will enter the water after 1 second.</p>Velocity physicshttps://physicsgurus.com/88209/cliff-diver-jumps-ledge-initial-upward-velocity-second-enters?show=88210#a88210Thu, 18 May 2023 10:03:25 +0000Answered: What will happen to the velocity and position of an object if we accelerate it with constant force?
https://physicsgurus.com/87815/happen-velocity-position-object-accelerate-constant-force?show=87816#a87816
<p>If an object is subjected to a constant force and undergoes acceleration, its velocity and position will change over time.</p><ol><li>Velocity: The object's velocity will increase over time in the direction of the applied force. The rate of change of velocity, or acceleration, is directly proportional to the applied force according to Newton's second law of motion:</li></ol><p>F = m * a</p><p>Where: F is the applied force m is the mass of the object a is the acceleration</p><p>If the force remains constant, the object's velocity will experience a steady increase as long as the force continues to act. The relationship between velocity, acceleration, and time can be described by the following equation:</p><p>v = u + a * t</p><p>Where: v is the final velocity u is the initial velocity a is the constant acceleration t is the time elapsed</p><ol start="2"><li>Position: The object's position will also change as it accelerates. The relationship between position, velocity, acceleration, and time can be described by the equations of motion. Assuming the object starts from an initial position (s_0), the position at any given time (t) can be calculated using the equation:</li></ol><p>s = s_0 + u * t + (1/2) * a * t^2</p><p>Where: s is the final position s_0 is the initial position u is the initial velocity a is the constant acceleration t is the time elapsed</p><p>As time progresses, both the object's velocity and position will change due to the constant force applied to it. The specific behavior of the object will depend on the magnitude and direction of the force, as well as its mass and any external factors affecting its motion.</p>Velocity physicshttps://physicsgurus.com/87815/happen-velocity-position-object-accelerate-constant-force?show=87816#a87816Thu, 18 May 2023 10:03:25 +0000Answered: Why does the vertical component of the velocity of a particle in projectile motion disappear once it reaches its maximum height?
https://physicsgurus.com/87627/vertical-component-velocity-particle-projectile-disappear?show=87628#a87628
<p>In projectile motion, the vertical component of the velocity of a particle decreases until it reaches its maximum height, at which point it momentarily becomes zero. This occurs because of the influence of gravity on the motion of the particle.</p><p>When a particle is launched into projectile motion, it moves along a curved path determined by its initial velocity and the gravitational force acting on it. The motion can be divided into two components: horizontal and vertical.</p><p>The horizontal component of velocity remains constant throughout the entire trajectory, assuming there are no external forces acting horizontally (such as air resistance). This is due to the absence of any force acting horizontally, resulting in a constant velocity.</p><p>The vertical component of velocity, on the other hand, is affected by the force of gravity. Gravity acts to accelerate the particle downward, causing its vertical velocity to decrease. As the particle moves upward, against the force of gravity, its vertical velocity gradually decreases until it reaches its maximum height.</p><p>At the maximum height, the vertical velocity becomes momentarily zero because the particle momentarily comes to a stop in the vertical direction. However, the horizontal velocity component continues to be constant, ensuring that the particle continues to move horizontally.</p><p>After reaching the maximum height, the particle begins to descend, and the force of gravity causes the vertical velocity to increase in the downward direction. This process continues until the particle returns to the ground.</p><p>In summary, the disappearance of the vertical component of velocity at the maximum height in projectile motion is a result of the deceleration due to gravity, causing the particle to momentarily come to a stop in the vertical direction before descending.</p>Velocity physicshttps://physicsgurus.com/87627/vertical-component-velocity-particle-projectile-disappear?show=87628#a87628Thu, 18 May 2023 10:03:25 +0000Answered: What is the velocity of a car which travels a distance of 50 kilometres in three hours?
https://physicsgurus.com/87327/what-velocity-which-travels-distance-kilometres-three-hours?show=87328#a87328
<p>To calculate the velocity of a car that travels a distance of 50 kilometers in three hours, we can use the formula:</p><p>Velocity = Distance / Time</p><p>Given: Distance = 50 kilometers Time = 3 hours</p><p>Substituting the given values into the formula:</p><p>Velocity = 50 km / 3 h ≈ 16.67 km/h</p><p>Therefore, the velocity of the car is approximately 16.67 kilometers per hour.</p>Velocity physicshttps://physicsgurus.com/87327/what-velocity-which-travels-distance-kilometres-three-hours?show=87328#a87328Thu, 18 May 2023 10:03:25 +0000What is the relationship between velocity and pressure? What is the relationship between velocity and force?
https://physicsgurus.com/88363/relationship-between-velocity-pressure-relationship-velocity
Velocity physicshttps://physicsgurus.com/88363/relationship-between-velocity-pressure-relationship-velocityWed, 17 May 2023 10:03:25 +0000