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arsen [322]
3 years ago
8

The diagram shows a ball resting at the top of a hill.

Physics
2 answers:
andrey2020 [161]3 years ago
8 0

Answer:

The potential energy in the system is greatest at X.

The first option

WARRIOR [948]3 years ago
7 0

Answer:

a) The potential energy in the system is greatest at X.

Explanation:

Let be X the point where a ball rests at the top of a hill. By applying the Principle of Energy Conservation, the total energy in the physical system remains constant and gravitational potential energy at the top of the hill is equal to the sum of kinetic energy, a lower gravitational energy and dissipated work due to nonconservative forces (friction, dragging).

U_{grav, X} = U_{grav, Y} + K_{Y} + \Delta W_{X \rightarrow Y} = U_{grav,Z}+\Delta W_{X \rightarrow Z}

Conclusions are showed as follows:

a) The potential energy in the system is greatest at X.

b) The kinetic energy is the lowest at X and Z.

c) Total energy remains constant as the ball moves from X to Y.

Hence, the correct answer is A.

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The minimum stopping distance of a car moving at 20.5 mi/h is 11.6 m. Under the same conditions (so that the maximum braking for
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Answer:

d = 69 .57 meter

Explanation:

First case

Speed of car ( v )  = 20.5 mi/h  = 9.164  M/S

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Work done = 0.5 m v²  = 0.5 * 9.164² * m J  = 41.99 m J

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3 years ago
If a moving car speeds up until it is going twice as fast, how much kinetic energy doe s it have compared with its initial kinet
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K_f =  \frac{1}{2} mv_f^2 =  \frac{1}{2}m (2 v_i)^2 = 4 ( \frac{1}{2} mv_i^2)=4 K_i
so, if the velocity of the car is doubled, the new kinetic energy is 4 times the initial kinetic energy.
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