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3 years ago

A ball with 100 J of PE is released from a height of 10 m. What will be the KE of the ball at 5

Physics

1 answer:

harkovskaia [24]3 years ago

8 0

Answer:

The kinetic energy is: 50[J]

Explanation:

The ball is having a potential energy of 100 [J], therefore

PE = [J]

The elevation is 10 [m], and at this point the ball is having only potential energy, the kinetic energy is zero.

$E_{p} =m*g*h\\where:\\g= gravity[m/s^{2} ]\\m = mass [kg]\\m= \frac{E_{p} }{g*h}\\ m= \frac{100}{9.81*10}\\\\m= 1.01[kg]\\\\$

In the moment when the ball starts to fall, it will lose potential energy and the potential energy will be transforme in kinetic energy.

When the elevation is 5 [m], we have a potential energy of

$P_{e} =m*g*h\\P_{e} =1.01*9.81*5\\\\P_{e} = 50 [J]\\$

This energy is equal to the kinetic energy, therefore

Ke= 50 [J]

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3 years ago

In a series RLC resonance circuit, the resonance frequency f0 = 700 kHz. The resistor R = 10 Ohm. The specified bandwidth (BW) s

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Answer:

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