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A body projected vertically from the earth reaches a height equal to earth’s radius before returning to the earth. The power exerted by the gravitational force is greatest
When a body is projected vertically from the Earth and reaches a height equal to the Earth's radius before returning, we can analyze the power exerted by the gravitational force throughout its motion. The gravitational force acts as a constant downward force, affecting the body's velocity as it riseRead more
When a body is projected vertically from the Earth and reaches a height equal to the Earth’s radius before returning, we can analyze the power exerted by the gravitational force throughout its motion. The gravitational force acts as a constant downward force, affecting the body’s velocity as it rises and falls.
Initially, at the moment just after the body is projected, it has its maximum velocity, resulting in a certain amount of kinetic energy. As the body ascends, gravity does negative work on it, causing its velocity to decrease until it reaches its maximum height, where the velocity becomes zero. At this point, the power exerted by the gravitational force is also zero, as power is the product of force and velocity.
As the body starts descending, it accelerates due to gravity, and its velocity increases. Just before the body hits the Earth, it reaches its maximum velocity once again. The gravitational force, although constant, does more work as the velocity of the body increases. Therefore, the power exerted by the gravitational force is greatest at the instant just before the body strikes the Earth, as both the gravitational force and the velocity are at their maximum values at that point.
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A small mass attached to string rotates on a frictionless table top as shown. If the tension in the string is increased by pulling the string causing the radius of the circular motion to decrease by a factor of 2, the kinetic energy of the mass will
To find how the kinetic energy of a mass changes when the radius of its circular motion is decreased by a factor of 2 while the tension in the string is increased, we can use the following relationships: 1. The centripetal force required to keep the mass moving in a circular path is provided by theRead more
To find how the kinetic energy of a mass changes when the radius of its circular motion is decreased by a factor of 2 while the tension in the string is increased, we can use the following relationships:
1. The centripetal force required to keep the mass moving in a circular path is provided by the tension in the string:
T = (m * v²) / r
where T is the tension, m is the mass, v is the velocity, and r is the radius.
2. The kinetic energy (K.E.) of the mass is given by:
K.E. = (1/2) * m * v².
3. If the radius is reduced to half of the previous value (r’ = r/2) and assuming tension is increased to a level at which circular motion will be maintained, we have
T’ = (m * v’²) / (r/2)
T’ = (2 * m * v’²) / r.
4. Keeping T’ constant and solving for v’, we see that v’ must be greater to be in equilibrium.
5. Putting it all back together to find the kinetic energy:
K.E.’ = (1/2) * m * (v’)².
Since v’ = √2 * v, the kinetic energy is multiplied by a factor of 4.
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Do all the electrons that absorb a photon come out as photoelectrons?
No, not all electrons that absorb a photon are ejected as photoelectrons. Only electrons that gain energy equal to or greater than the material's work function can escape. Others lose energy through collisions or remain bound within the material. For more visit here: https://www.tiwariacademy.com/ncRead more
No, not all electrons that absorb a photon are ejected as photoelectrons. Only electrons that gain energy equal to or greater than the material’s work function can escape. Others lose energy through collisions or remain bound within the material.
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A body of mass 1 kg is thrown upwards with a velocity of 20 m/s. It momentarily comes to rest after attaining a height of 18 m. How much energy is lost due to air friction? [Take g = 10 m/s²]
To find the energy lost due to air friction, we first calculate the initial kinetic energy and the potential energy at the maximum height. 1. Initial kinetic energy (K.E.) when the body is thrown: K.E. = (1/2) * m * v² K.E. = (1/2) * 1 kg * (20 m/s)² K.E. = (1/2) * 1 * 400 K.E. = 200 J. 2. PotentialRead more
To find the energy lost due to air friction, we first calculate the initial kinetic energy and the potential energy at the maximum height.
1. Initial kinetic energy (K.E.) when the body is thrown:
K.E. = (1/2) * m * v²
K.E. = (1/2) * 1 kg * (20 m/s)²
K.E. = (1/2) * 1 * 400
K.E. = 200 J.
2. Potential energy (P.E.) at the maximum height (h = 18 m):
P.E. = m * g * h
P.E. = 1 kg * 10 m/s² * 18 m
P.E. = 180 J.
3. The energy lost to air friction is given by the difference between the initial kinetic energy and the potential energy at the maximum height:
See lessEnergy lost = K.E at the start – P.E.
Energy lost = 200 J – 180 J
Energy lost = 20 J.
The coefficient of restitution e for a perfectly elastic collision is
The coefficient of restitution, denoted by 'e', describes how elastic the collision between two bodies is. When the two bodies collide and go back into their original shape without any changes in momentum or kinetic energy after a perfectly elastic collision, it follows that, by definition of e, eRead more
The coefficient of restitution, denoted by ‘e’, describes how elastic the collision between two bodies is. When the two bodies collide and go back into their original shape without any changes in momentum or kinetic energy after a perfectly elastic collision, it follows that, by definition of e,
e=relative velocity after collision/ relative velocity before collision
For the perfectly elastic collision, e equals 1.
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