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elena55 [62]
3 years ago
8

The free-fall acceleration at the surface of planet 1 is 26 m/s^2 . The radius and the mass of planet 2 are twice those of plane

t 1.
What is the free-fall acceleration on planet 2? Express your answer using two significant figures. m/s^2.
Physics
1 answer:
valkas [14]3 years ago
6 0

Answer:

13 m/s^2

Explanation:

The acceleration of gravity near the surface of a planet is:

g = MG / R^2

For planet 1, g = 26 m/s^2.

The gravity on planet 2 in terms of the mass and radius of planet 1 is:

g = (2M)G / (2R^2)

g = 1/2 MG / R^2

Since MG/R^2 = 26 m/s^2, then:

g = 13 m/s^2

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Calculate the internal axial load at a point D if length L=7 ft. The part is subjected to loads P1=632 lbs, P2=888 lbs (applied
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Answer:

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3 years ago
What is the proportionality between pressure and temperature, and the proportionality between atmospheric pressure and tempratur
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6 0
3 years ago
A particle is moving along a projectile path where the initial height is 160 feet with an initial speed of 144 feet per second.
Leviafan [203]
Assuming Earth's gravity, the formula for the flight of the particle is: 

<span>s(t) = -16t^2 + vt + s = -16t^2 + 144t + 160. </span>

<span>This has a maximum when t = -b/(2a) = -144/[2(-16)] = -144/(-32) = 9/2. </span>

<span>Therefore, the maximum height is s(9/2) = -16(9/2)^2 + 144(9/2) + 160 = 484 feet. </span>
7 0
3 years ago
Read 2 more answers
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