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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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The amplitude of a standing sound wave in a long pipe closed at the left end is sketched below.
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(1) The harmonic number for the mode of oscillation is 3.

(2) The pitch (frequency) of the sound is 579.55 Hz

(3) The level of the water inside the vertical pipe is 0.1 m.

<h3>The harmonic number</h3>

The harmonic number for the mode of oscillation illustrated for the closed pipe is 3.

<h3>Frequency of the wave</h3>

The pitch (frequency) of the sound is calculated from third harmonic formula;

f = 3v/4L

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f = (3 x 340) / (4 x 0.44)

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<h3>level of the water</h3>

wave equation for first harmonic of a closed pipe is given as

f  = v/(4L)

251.1  = 340/(4L)

4L = 340/251.1

4L = 1.35

L = 1.35/4

L = 0.34 m

level of water = 0.44 m - 0.34 m = 0.1 m

Thus, the level of the water inside the vertical pipe is 0.1 m.

Learn more about harmonics of closed pipes here: brainly.com/question/27248821

#SPJ1

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