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asambeis [7]
3 years ago
12

Find the magnitude of the free-fall acceleration at the orbit of the Moon (a distance of 60RE from the center of the Earth with

the radius of RE). Use the gravitational acceleration on the surface of the Earth is 9.8m/s2.
Physics
1 answer:
Ede4ka [16]3 years ago
6 0

Answer:

The magnitude of the free-fall acceleration at the orbit of the Moon is 2.728\times 10^{-3}\,\frac{m}{s^{2}} (\frac{2.784}{10000}\cdot g, where g = 9.8\,\frac{m}{s^{2}}).

Explanation:

According to the Newton's Law of Gravitation, free fall acceleration (g), in meters per square second, is directly proportional to the mass of the Earth (M), in kilograms, and inversely proportional to the distance from the center of the Earth (r), in meters:

g = \frac{G\cdot M}{r^{2}} (1)

Where:

G - Gravitational constant, in cubic meters per kilogram-square second.

M - Mass of the Earth, in kilograms.

r - Distance from the center of the Earth, in meters.

If we know that G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}, M = 5.972\times 10^{24}\,kg and r = 382.26\times 10^{6}\,m, then the free-fall acceleration at the orbit of the Moon is:

g = \frac{\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (5.972\times 10^{24}\,kg)}{(382.26\times 10^{6}\,m)^{2}}

g = 2.728\times 10^{-3}\,\frac{m}{s^{2}}

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During a plane showcase, a pilot makes circular "looping" with a speed equal to the sound speed (340 m/s). However, the pilot ca
Dmitriy789 [7]

Answer:

1472.98 m

Explanation:

Data provided:

Speed of circular looping, v = 340 m/s

Acceleration, a = 8g

here,

g is the acceleration due to the gravity = 9.81 m/s²

Now,

the centripetal acceleration is given as,

a=\frac{v^2}{r}

r is the radius of the loop

on substituting the respective values, we get

8\times9.81=\frac{340^2}{r}

or

r = 1472.98 m

5 0
4 years ago
The diagram below shows different weights on the see-saw. Will the see-saw move?
oee [108]

Answer:

yes it will

Explanation:

8 0
3 years ago
A 50 g mass is freely hanging from a horizontal meter stick at a distance of 99 cm from the pivot. Calculate the weight force W
Neko [114]

Answer:

W = 0.49 N

τ = 0.4851 Nm

Force

Explanation:

The weight force can be found as:

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W = (0.05 kg)(9.8 m/s²)

<u>W = 0.49 N</u>

The torque about the pivot can be found as:

τ = W*d

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τ = torque

d = distance between weight and pivot = 99 cm = 0.99 m

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τ = (0.49 N)(0.99 m)

<u>τ = 0.4851 Nm</u>

The pivot exerts a  <u>FORCE </u>on the meter stick because the pivot applies force normally over the stick and has a zero distance from stick.

6 0
3 years ago
A cart travels 4.00 meters east and then4.00 meters north. determine the magnitude ofthe cart’s resultant displacement.
gogolik [260]
^
|
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Square root of (4^2 + 4^2) = 4*squareRoot(2)
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tensa zangetsu [6.8K]
The answer for question 2 i guess it’s c
3 0
2 years ago
Read 2 more answers
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