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Serhud [2]
4 years ago
7

Suppose a ball has a potential energy of 5 J when you drop it. What would be its kinetic energy just as it hit the ground? (Igno

re the effect of air resistance.)
Physics
1 answer:
mihalych1998 [28]4 years ago
8 0

If you simply drop it ... so that it's not moving until you let it go ... then
the kinetic energy it has when it hits the ground is exactly the potential
energy it had while you were holding it.

5 J potential energy in your hand  ===>  5 J kinetic energy when it lands


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velocity = 62.89 m/s  in 58 degree measured from the x-axis

Explanation:

Relevant information:

Before the collision, asteroid A of mass 1,000 kg moved at 100 m/s, and asteroid B of mass 2,000 kg moved at 80 m/s.

Two asteroids moving with velocities collide at right angles and stick together. Asteroid A initially moving to right direction and asteroid B initially move in the upward direction.

Before collision Momentum of A = 1000 x 100 = $ 10^5$ kg - m/s in the right direction.

Before collision Momentum of B = 2000 x 80 = 1.6 x $ 10^5$  kg - m/s in upward direction.

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Initial momentum in right direction = final momentum in right direction = $ 10^5$

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and $ V_y=\frac{160}{3}$  m/s

Therefore, velocity is = $ \sqrt{V_x^2 + V_y^2} $

                                   = $ \sqrt{(\frac{100}{3})^2 + (\frac{160}{3})^2} $

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And direction is

tan θ = $ \frac{V_y}{V_x}$     = 1.6

therefore, $ \theta = \tan^{-1}1.6 $

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3 years ago
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