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sergij07 [2.7K]
4 years ago
13

The Moon is about 3.8 ✕ 108 m from Earth. Traveling at the speed of light, 3.0 ✕ 108 m/s, how long does it take a laser beam to

go from Earth to the Moon and back again (in s)? The same physics was responsible for the noticeable delay in communications signals between lunar astronauts and controllers at the Houston Space Flight Center
Physics
1 answer:
lana66690 [7]4 years ago
4 0
Hello

1) The total distance the light has to cover to go from Earth to the Moon and back is twice the distance between the Earth and the Moon:
S=2d=2\cdot 3.8 \cdot 10^8~m =7.6 \cdot 10^8~m

2) Given the relationship
v= \frac{S}{t}
where v=3\cdot10^8~m/s is the speed of light, S the total distance covered calculate at point 1), and t the time, we can find the time the light needs to go from Earth to the Moon and back again:
t= \frac{S}{v}= \frac{7.6\cdot10^8 m}{3\cdot10^8~m/s}=2.53~s
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A baseball has a mass of 0.15 kg and radius 3.7 cm. In a baseball game, a pitcher throws the ball with a substantial spin so tha
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Answer:

Rotational kinetic energy = 0.099 J

Translational kinetic energy = 200 J

The moment of inertia of a solid sphere is I = \frac{2}{5}mr^2.

Explanation:

Rotational kinetic energy is given by

\text{RKE} = \frac{1}{2}I\omega^2

where <em>I</em> is the moment of inertia and <em>ω</em> is the angular speed.

For a solid sphere,

I = \frac{2}{5}mr^2

where <em>m</em> is its mass and <em>r</em> is its radius.

From the question,

<em>ω</em> = 49 rad/s

<em>m</em> = 0.15 kg

<em>r</em> = 3.7 cm = 0.037 m

\text{RKE} = \frac{1}{2}\times \frac{2}{5} mr^2\omega^2 = \frac{1}{5} mr^2\omega^2

\text{RKE} = \frac{1}{5} (0.15\ \text{kg})(0.037\ \text{m})^2(49\ \text{rad/s})^2 = 0.099\text{ J}

Translational kinetic energy is given by

\text{TKE} = \frac{1}{2} mv^2

where <em>v</em> is the linear speed.

\text{TKE} = \frac{1}{2} (0.15\ \text{kg})(52\ \text{m/s})^2 = 200\text{ J}

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