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DerKrebs [107]
3 years ago
11

A camera rests on a platform that is 10 meters above the ground. what kind of energy does it possess?

Physics
2 answers:
Delvig [45]3 years ago
8 0

Answer:

this is gravitational potential energy that is stored in that position of camera

Explanation:

As we know that the camera is at height 10 m from the ground

h = 10 m

also we know that camera is at the position of 10 m from the ground at rest position

So here we can say that kinetic energy of the camera is ZERO

so here the energy is possessed due to the position of the camera and it is the gravitational potential energy that is stored in it

So here the energy is given as

U = mgH

so this is gravitational potential energy that is stored in that position of camera

victus00 [196]3 years ago
7 0
Is the camera moving?
Nope, it's at rest.
No kinetic energy there.

Is it at some distance above ground?
Yepp, height of 10m.
You've got gravitational potential energy!

To calculate it you'll use the equation:

energy = mgh
... if you want to. ;)
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F_B=G\frac{m_A m_B}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(567 kg)}{(0.500 m)^2}=5.14 \cdot 10^{-5} N to the right

The gravitational force exerted by particle C on particle A is

F_C=G\frac{m_A m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(139 kg)}{(0.500 m+0.250m)^2}=5.6 \cdot 10^{-6} N to the right

So the net gravitational force on particle A is

F_A = F_B + F_C =5.14 \cdot 10^{-5} N+5.6 \cdot 10^{-6} N=5.7 \cdot 10^{-5} N to the right


(b) Particle B

The gravitational force exerted by particle A on particle B is

F_A=G\frac{m_A m_B}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(567 kg)}{(0.500 m)^2}=-5.14 \cdot 10^{-5} N to the left

The gravitational force exerted by particle C on particle B is

F_C=G\frac{m_B m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(567 kg)(139 kg)}{(0.250 m)^2}=8.41 \cdot 10^{-5} N to the right

So the net gravitational force on particle B is

F_B = F_A + F_C =-5.14 \cdot 10^{-5} N+8.41 \cdot 10^{-5} N=3.27 \cdot 10^{-5} N to the right


(c) Particle C

The gravitational force exerted by particle A on particle C is

F_A=G\frac{m_A m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(139 kg)}{(0.500 m+0.250m)^2}=-5.6 \cdot 10^{-6} N to the left

The gravitational force exerted by particle B on particle C is

F_B=G\frac{m_B m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(567 kg)(139 kg)}{(0.250 m)^2}=-8.41 \cdot 10^{-5} N to the left

So the net gravitational force on particle C is

F_C = F_B + F_A =-8.41 \cdot 10^{-5} N-5.6 \cdot 10^{-6} N=-8.97 \cdot 10^{-5} N to the left



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