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Ronch [10]
3 years ago
11

How can you increase the potential energy of a bouncing ball

Physics
1 answer:
ad-work [718]3 years ago
7 0

Answer:

When you lift the ball, you are doing work to increase its gravitational potential energy. When you then release the ball, gravitational energy is transformed into kinetic energy as the ball falls. When the ball hits the floor, the ball's shape changes as it flattens against the floor.

Explanation:thats should be the way^^ in explaining

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An object is hanging by a string from the ceiling of an elevator. the elevator is moving upward with a constant speed. what is t
Anarel [89]

Since the elevator is moving with a constant speed and not accelerating, the tension in the string is simply the normal, routine, everyday boring weight of the object.  Since the elevator is moving with a constant speed and not accelerating, the tension in the string is simply the normal, routine, everyday boring weight of the object.  

6 0
4 years ago
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The image shows one complete cycle of a mass on a spring in simple harmonic motion. An illustration of a mass on a vertical spri
Alja [10]

Answer:

D. "The net force is zero, so the acceleration is zero"

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6 0
3 years ago
How to find the coefficient of kinetic friction on an incline?
PolarNik [594]

Answer:

An object slides down an inclined plane at a constant velocity if the net force on the object is zero. We can use this fact to measure the coefficient of kinetic friction between two objects.

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7 0
3 years ago
I need help with #10
kherson [118]
Conduction. Because they are connected as they tranfer energy
3 0
3 years ago
A basketball player makes a jump shot. The 0.599 kg ball is released at a height of 2.18 m above the floor with a speed of 7.05
7nadin3 [17]

Answer:

W_{drag} = 4.223\,J

Explanation:

The situation can be described by the Principle of Energy Conservation and the Work-Energy Theorem:

U_{g,A}+K_{A} = U_{g,B} + K_{B} + W_{drag}

The work done on the ball due to drag is:

W_{drag} = (U_{g,A}-U_{g,B})+(K_{A}-K_{B})

W_{drag} = m\cdot g\cdot (h_{A}-h_{B})+ \frac{1}{2}\cdot m \cdot (v_{A}^{2}-v_{B}^{2})

W_{drag} = (0.599\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (2.18\,m-3.10\,m)+\frac{1}{2}\cdot (0.599\,kg)\cdot [(7.05\,\frac{m}{s} )^{2}-(4.19\,\frac{m}{s} )^{2}]

W_{drag} = 4.223\,J

7 0
4 years ago
Read 2 more answers
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