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VMariaS [17]
1 year ago
10

The moon has a mass of 1×10^22 kg, and the gravitational field strength at a distance r from the planet is 0.001 n/kg. what is t

he gravitational force exerted on the moon while it is in orbit around the planet?
Physics
1 answer:
CaHeK987 [17]1 year ago
4 0

When I divided by 6, I got W(moon) = mg(moon) = 10 x 10/6 = 16.7N (the moon's gravity is only 1/6 that of Earth in this case). 100N is equal to W(Earth) = mg(Earth) = 10 x 10. On the moon and the Earth, m = 10 kg.

How can you determine the strength of Earth's gravitational field at the moon?

M = 6 10 24 K g is the mass of the earth, and g = - G M R 2 is the gravitational field of the earth at the point of the moon. The distance between the earth and moon is 84 10 8 meters. field of gravitation.

How do you determine the distance between Earth and the moon where the gravitational field is at its weakest?

You only discover points where the forces from all of the different local bodies are in balance, or equal to each other, because there is no place in the universe where the gravitational field intensity is zero. All bodies, not just the Earth and Moon, must be taken into account in the computation to arrive at these positions.

To know more about gravitational field strength at moon visit;

brainly.com/question/20909424

#SPJ4

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Four weightlifters (A-D) enter a competition. The mass, distance, and time of their lifts are shown in the table.
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Let Pa, Pb, Pc, and Pd be the powers delivered by weightlifters A, B, C, and D, respectively.

Use this equation to determine each power value:

P = W÷Δt

P is the power, W is the work done by the weightlifter, and Δt is the elapsed time.

A) Determining Pa:

Pa = W÷Δt

The weightlifter does work to lift the weight up a certain distance. Therefore the work done is equal to the weight's gain in gravitational potential energy. The equation for gravitational PE is

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PE is the potential energy, m is the mass of the weight, g is the acceleration of objects due to earth's gravity, and h is the distance the weight was lifted.

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Plug in the values and solve for Pa

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B) Determining Pb:

Let us use our new equation derived in part A to solve for Pb:

Pb = mgh÷Δt

Given values:

m = 150.0kg

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Plug in the values and solve for Pb

Pb = 150.0×9.81×1.76÷0.052

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C) Determining Pc:

Pc = mgh÷Δt

Given values:

m = 200.0kg

g = 9.81m/s²

h = 1.50m

Δt = 0.217s

Plug in the values and solve for Pc

Pc = 200.0×9.81×1.50÷0.217

<u>Pc = 13600W</u>

D) Determining Pd:

Pd = mgh÷Δt

Given values:

m = 250.0kg

g = 9.81m/s²

h = 1.25m

Δt = 0.206s

Plug in the values and solve for Pd

Pd = 250.0×9.81×1.25÷0.206

<u>Pd = 14900W</u>

Compare the following power values:

Pa = 14600W, Pb = 49800W, Pc = 13600W, Pd = 14900W

Pc is the lowest value.

Therefore, weightlifter C delivers the least power.

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