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s2008m [1.1K]
4 years ago
6

When you throw a ball upward, its kinetic energy ____ (DECREASES,INCREASES,REMAINS THE SAME) and its potential energy _____ (DEC

REASE,INCREASES, REMAINS THE SAME) . When the ball reaches maximum height, its kinetic energy is at a ____ (maximum, minimum).
Physics
2 answers:
4vir4ik [10]4 years ago
7 0

When you throw a ball upward, its kinetic energy decreases and its potential energy increases. When the ball reaches maximum height, its kinetic energy is at a minimum.

Explanation:

- When the ball is thrown upward, its kinetic energy decreases. In fact, the kinetic energy of the ball is given by

K=\frac{1}{2}mv^2

where m is the mass of the ball and v its speed. We see that the kinetic energy is directly proportional to the speed: since the speed of the ball decreases as the ball gets higher, its kinetic energy decreases as well.

- the potential energy increases. In fact, the potential energy of the ball is given by

U=mgh

where g is the gravitational acceleration and h is the height above the ground. We see that U is directly proportional to h, therefore as the ball gets higher, h increases and the potential energy increases as well.

- When the ball reaches maximum height, its kinetic energy is minimum. To be more precise, the kinetic energy becomes zero. In fact, at the point of maximum height the speed of the ball becomes zero (because the ball changes direction, it starts going downward), so v=0 in the formula of the kinetic energy and therefore the kinetic energy is zero.

Anettt [7]4 years ago
6 0
When the ball goes upward, the velocity is decreasing. Therefore, the kinetic energy is decreased. The potential energy when the ball's height increases also increases. When the ball reaches maximum, the velocity becomes zero and the kinetic energy is minimum. 
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Answer:

27 m/s

Explanation:

Given:

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Explanation:

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An ice sled powered by a rocket engine starts from rest on a large frozen lake and accelerates at 12.8 m/s2 . At t1 the rocket e
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Explanation:

Given

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acceleration a=12.8 m/s^2

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v=0+at

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distance traveled by sled in t_1 s

v^2-u^2=2as

(12.8t_1)^2-0=2\times 12.8\times s_1

s_1=6.4\cdot t_1^2

distance traveled in t_2 time with velocity v=12.8t_1

s_2=v\times t_2

s_2=12.8\times t_1\times t_2

s_2=12.8\cdot t_1\cdot t_2

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t_1+t_2=97.7 s

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Answer:

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Explanation:

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strojnjashka [21]

Answer:

The correct answer is t = 0.92s

Explanation:

Initial velocity v​0 = 3.0 m/s

Displacement Δy = ?

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-9.8m/s2

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