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Fed [463]
3 years ago
9

The masses of the Moon and Earth are 7.35 x 1022 kg and 5.97 x 1024 kg, respectively. The strength of the gravitational force be

tween them is 1.98 x 1020 N. What is distance between the centre of the Earth and the centre of the Moon?
Physics
1 answer:
bixtya [17]3 years ago
3 0

Answer:

Distance between centre of Earth and centre of Moon is 3.85 x 10⁸ m

Explanation:

The attractive force experienced by two mass objects is known as Gravitational force.

The gravitational force is determine by the relation:

F=\frac{Gm_{1} m_{2} }{d^{2} }      ....(1)

According to the problem,

Mass of Moon, m₁ = 7.35 x 10²² kg

Mass of Earth, m₂ = 5.97 x 10²⁴ kg

Gravitational force experienced by them, F = 1.98 x 10²⁰ N

Universal gravitational constant, G = 6.67 x 10⁻¹¹ Nm²kg⁻²

Substitute these values in equation (1).

1.98\times10^{20} =\frac{6.67\times10^{-11}\times7.35\times10^{22}\times5.97\times10^{24} }{d^{2} }

d^{2}=\frac{2.93\times10^{37}}{1.98\times10^{20}}

d=\sqrt{1.48\times10^{17}}

d = 3.85 x 10⁸ m

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Answer:

The amount of electrons that flow in the given time is 3.0 C.

Explanation:

An electric current is defined as the ratio of the quantity of charge flowing through a conductor to the time taken.

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It is measure in Amperes and can be measured in the laboratory by the use of an ammeter.

In the given question, I = 1.5A, t = 2s, find Q.

From equation 1,

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               = 3.0 Coulombs

The amount of electrons that flow in the given time is 3.0 C.

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1 year ago
Why is it important for scientist to do research?
sweet-ann [11.9K]

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3 years ago
Read 2 more answers
Holding onto a tow rope moving parallel to a frictionless ski slope, a 70.1 kg skier is pulled up the slope, which is at an angl
lara [203]

Answer:

given,

mass of the skier = 70.1 Kg

angle with horizontal, θ = 8.6°

magnitude of the force,F = ?

a) Applying newton's second law

   m g sin\theta - F_{rope} = ma

 velocity is constant, a = 0

   m g sin\theta - F_{rope} =0

   F_{rope} = m g sin\theta

   F_{rope} = 70.1\times 9.8\times sin 8.6^0

  F_{rope}= 102.73\ N

b) now, when acceleration, a = 0.135 m/s²

   F_{rope}-m g sin\theta = ma

 velocity is constant, a = 0.135 m/s₂

   F_{rope} = m g sin\theta+ma

   F_{rope} = 70.1\times 9.8\times sin 8.6^0+70.1\times 0.135

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7 0
3 years ago
If a marble is released from a height of 10 meters how long would it take to hit the ground?
zlopas [31]
S = ut + 0.5at^2 

<span>10 = 0 + 0.5(9.81)t^2 {and if g = 10 then t^2 = 2 so t ~1.414}  </span>

<span>t^2 ~ 2.04 </span>

<span>t ~ 1.43 seconds</span>
3 0
3 years ago
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