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suter [353]
3 years ago
8

What is the gravitational acceleration close to the surface of a planet with a mass of 9ME and radius of 3RE, where ME and RE ar

e the mass and radius of Earth, respectively?
Physics
1 answer:
Papessa [141]3 years ago
5 0

Answer:

9.78 m/s²

Explanation:

To solve this, we use the gravitational formula

g = GM/r², where

g = acceleration due to gravity

G = gravitational constant

M = mass of the planet

r = radius of the planet

From the question, we got that the mass of the planet is

M = 9ME, where ME = 5.95*10^24

M = 9 * 5.95*10^24

M = 5.355*10^25 kg

Also, the Radius of the planet, R = 3RE, where RE = 6.37*10^6

R = 3 * 6.37*10^6

R = 1.911*10^7 m

On applying the values of both R and M to the equation, we get

g = GM/r²

g = (6.67*10^-11 * 5.355*10^25) / (1.911*10^7)²

g = 3.57*10^15/3.65*10^14

g = 9.78 m/s²

Therefore, the acceleration due to gravity on the planet is 9.78 m/s²

Please vote brainliest if it helped you <3

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Amanda [17]

Answer:

S = V0 t + 1/2 a t^2

S = 5 m/s * 300 s + 1/2 * 1.2 m/s * (300 s^2)

S = 1500 m + .6 * 90000 m = 55,500 m

Check:     V0 = 5 m/s

                V2 = V0 + a t  = 5 + 1.2 * 300 = 365 m/s

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S = 185 * 300 = 55,500 m

We calculated V2 above at 365 m/s  the speed after 300 sec

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2 years ago
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antoniya [11.8K]

Answer:

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Explanation:

In the metric system, the standard units for the below are;

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8 0
3 years ago
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____ [38]
The correct answer is "As the distance from the earth increases, the gravitational pull on the spaceship would decrease."

In fact, the gravitational force (attractive) exerted by the Earth on the spaceship is given by
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