2 years ago

At any given time in space, how much of the moon is lit by the sun? PLEASE ANSWER QUICKLY!!!

1 answer:

3 0

Answer: 50%

Explanation: The sun and the moon go around Earth and The sun is way bigger than the moon so 50% of the sun lights up the moon at night

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If you are pushing on a box with a force of 20 N and there is a 7 N force on the box due to sliding friction, what is the net fo

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Answer:

13 N

Explanation:

The Net Force of an object should be the difference between the forces applied to the object if the object is not in equilibrium. This object is not in equilibrium so therefore by finding the difference between the forces, you will find your answer. 20 N - 7 N = 13 N.

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A 1.5 kilogram car is moving at 10 meters per second east. A braking force acts on the car for 5.0 seconds, reducing its velocit

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Answer:

I don't know

Explanation:

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3 years ago

What is the average velocity of a car if it travels from position 25m to a position of -7m in 34 seconds?

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Answer:

vp = 0.94 m/s

Explanation

Formula

Vp = position/ time

position: Initial position - Final position

Position = 25 m - (-7 m) = 25 m + 7 m = 32 m

Then

Vp = 32 m / 34 seconds

Vp = 0.94 m/s

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Using energy considerations, calculate the average force (in N) a 67.0 kg sprinter exerts backward on the track to accelerate fr

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Answer:

$F_{sprinter}=110.4N$

Explanation:

Given data

Mass m=67.0 kg

Final Speed vf=8.00 m/s

Initial Speed vi=2.00 m/s

Distance d=25.0 m

Force F=30.0 N

From work-energy theorem we know that the work done equals the change in kinetic energy

W=ΔK=Kf-Ki=1/2mvf²-1/2mvi²

And

$W=F_{total}.d$

$W=1/2mv_{f}^2-1/2mv_{i}^2\\F_{total}=\frac{1/2mv_{f}^2-1/2mv_{i}^2}{d} \\F_{total}=\frac{1/2(67.0kg)(8.00m/s)^2-1/2(67.0kg)(2.00m/s)^2}{25.0m} \\F_{total}=80.4N$

and we know that the force the sprinter exerted Fsprinter the force of the headwind Fwind=30.0N

$F_{sprinter}=F_{total}+F_{wind}\\F_{sprinter}=80.4N+30N\\F_{sprinter}=110.4N$

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2 years ago

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Six keplerian element table

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