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Pavel [41]
3 years ago
13

Calculate the magnitude of the acceleration due to gravity on the surface of Earth due to the Moon.

Physics
1 answer:
Fudgin [204]3 years ago
4 0

Answer:

g'_h=1.096\times 10^{-5}\ m.s^{-2}

Explanation:

We know that the gravity on the surface of the moon is,

  • g'=\frac{g}{6}
  • g'=1.63\ m.s^{-2}

<u>Gravity at a height h above the surface of the moon will be given as:</u>

g'_h=\frac{G.m}{(r+h)^2} ..........................(1)

where:

G = universal gravitational constant

m = mass of the moon

r = radius of moon

We have:

  • G=6.67\times 10^{-11}\ m^3.s^{-2}.kg^{-1}
  • m=7.35\times 10^{22}\ kg
  • r=1.74\times 10^6\ m
  • h=384.4\times 10^6\ m is the distance between the surface of the earth and the moon.

Now put the respective values in eq. (1)

g'_h=\frac{6.67\times 10^{-11}\times 7.35\times 10^{22}}{(1.74\times 10^6+384.4\times 10^6)^2}

g'_h=1.096\times 10^{-5}\ m.s^{-2} is the gravity on the moon the earth-surface.

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Suppose the entire solar nebula had cooled to 50 K before the solar wind cleared the early solar system of its gases. How would the composition and sizes of the planets of the inner solar system be different from what we see today is given below

Explanation:

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4 0
3 years ago
A shopper walks westward 2.8 meters in the first 5 minutes and then eastward 9.2 meters in the next 10 minutes. What is the shop
Alecsey [184]

Answer:

12m

Explanation:

Given parameters:

Distance walked westward  = 2.8m

Time of travel  = 5min

Distance walked eastward = 9.2m

Time of travel  = 10min

Unknown:

The total shopper's travel distance  = ?

Solution:

Total distance traveled is the sum of the length of path covered by a body. It is a scalar quantity.

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3 years ago
A force causes a mass of 4 kg to have an acceleration of 8 m/s2. Suppose something causes the mass to be one-quarter of its orig
solniwko [45]

Explanation:

From Newton's second law:

F = ma

Given that m = 4 kg and a = 8 m/s²:

F = (4 kg) (8 m/s²)

F = 32 N

If m is reduced to 1 kg and F stays at 32 N:

32 N = (1 kg) a

a = 32 m/s²

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3 years ago
Interactive LearningWare 10.1 reviews the concepts involved in this problem. A spring stretches by 0.0161 m when a 3.74-kg objec
yKpoI14uk [10]

Answer:

m = 3.91 kg

Explanation:

Given that,

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mg=kx

k=\dfrac{mg}{x}

k=\dfrac{3.74\times 9.8}{0.0161}

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The frequency of vibration is given by :

f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}

m=\dfrac{k}{4\pi^2f^2}

m=\dfrac{2276.52}{4\pi^2\times (3.84)^2}

m = 3.91 kg

So, the mass of the object is 3.91 kg. Hence, this is the required solution.

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