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Georgia [21]
3 years ago
13

What is the potential energy of a 5kg rock that is 7m high on a hill

Physics
1 answer:
Goryan [66]3 years ago
4 0

Explanation:

Gravitational potential energy = mgh = (5)(9.81)(7) = 343.35J.

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No links please, links never work, this is science not physics
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You kick a ball with a speed of 14 m/s at an angle of 51°. How far away does the ball land?
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-- The vertical component of the ball's velocity is 14 sin(<span>51°) = 10.88 m/s

-- The acceleration of gravity is 9.8 m/s².

-- The ball rises for 10.88/9.8 seconds, then stops rising, and drops for the
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-- Altogether, the ball is in the air for (2 x 10.88)/(9.8) = 2.22 seconds
==================================

-- The horizontal component of the ball's velocity is  14 cos(</span><span>51°) = 8.81 m/s

-- At this speed, it covers a horizontal distance of (8.81) x (2.22) = <em><u>19.56 meters</u></em>
before it hits the ground.


As usual when we're discussing this stuff, we completely ignore air resistance.
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4 0
3 years ago
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Lightning produces a maximum air temperature on the order of 104K, whereas a nuclear explosion produces a temperature on the ord
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Answer:

tex]2.898\times 10^{-7}\ \text{m}[/tex] ultraviolet region

2.898\times 10^{-10}\ \text{m} x-ray region

Explanation:

T = Temperature

b = Constant of proportionality = 2.898\times 10^{-3}\ \text{m K}

\lambda = Wavelength

T=10^4\ \text{K}

From Wein's law we have

\lambda=\dfrac{b}{T}\\\Rightarrow \lambda=\dfrac{2.898\times 10^{-3}}{10^4}\\\Rightarrow \lambda=2.898\times 10^{-7}\ \text{m}

The wavelength of the radiation will be 2.898\times 10^{-7}\ \text{m} and it is in the ultraviolet region.

T=10^7\ \text{K}

\lambda=\dfrac{2.898\times 10^{-3}}{10^7}\\\Rightarrow \lambda=2.898\times 10^{-10}\ \text{m}

The wavelength of the radiation will be 2.898\times 10^{-10}\ \text{m} and it is in the x-ray region.

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