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AURORKA [14]
2 years ago
15

An astronaut with a mass of 91 kg is 0.30 m above the moons surface. The astronauts potential energy is 46 J. Calculate the free

-fall acceleration on the moon?
Physics
1 answer:
Blababa [14]2 years ago
3 0

Answer:

the free-fall acceleration on the moon is 1.68 m/s^2

Explanation:

recall the formula for the gravitational potential energy (under acceleration of gravity "g"):

PE = m * g * h

replacing with our values for the problem:

46 J = 91 * g * 0.3

solve for the "g" on the Moon:

g = 46 / (91 * 0.3)

g = 1.68  m/s^2

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What minimum distance should you separate two sources emitting the same waves with wavelength 5mm in phase such that you obtain
Makovka662 [10]

To solve this problem we will apply the concept related to destructive interference (from the principle of superposition). This concept is understood as a superposition of two or more waves of identical or similar frequency that, when interfering, create a new wave pattern of less intensity (amplitude) at a point called a node. Mathematically it can be described as

d = n \frac{\lambda}{2}

Where,

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d = \frac{\lambda}{2}

d = \frac{5*10^{-3}}{2}

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8 0
3 years ago
At what speed, as a fraction of c , is a particle's total energy twice its rest energy
WINSTONCH [101]
The rest energy of a particle is
E_0=m_0 c^2
where m_0 is the rest mass of the particle and c is the speed of light.

The total energy of a relativistic particle is
E=mc^2 =  \frac{m_0 c^2}{ \sqrt{1- \frac{v^2}{c^2} } }
where v is the speed of the particle.

We want the total energy of the particle to be twice its rest energy, so that
E=2E_0
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\frac{m_0c^2}{ \sqrt{1- \frac{v^2}{c^2} } }=2m_0 c^2
\frac{1}{ \sqrt{1- \frac{v^2}{c^2} } }=2
From which we find the ratio between the speed of the particle v and the speed of light c:
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So, the particle should travel at 0.87c in order to have its total energy equal to twice its rest energy.
3 0
3 years ago
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