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3 years ago

Captain Kirk (80.0 kg) beams down to a planet that is the same size as Uranus and finds that he weighs

Physics

1 answer:

Archy [21]3 years ago

8 0

The mass of the planet is $1.51\cdot 10^{26}kg$

Explanation:

The weight of Captain Kirk on the surface of the planet is equal to the gravitational force between him and the planet, which is:

$F=G\frac{Mm}{R^2}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

M is the mass of the planet

m is the mass of Captain Kirk

R is the radius of the planet

In this problem, we have:

m = 80.0 kg is the mass of Kirk

$R=25,362 km = 2.54\cdot 10^7 m$ is the radius of the planet (same as Uranus)

F = 1250 N is the magnitude of the gravitational force between Kirk and the planet

Solving for M, we find the mass of the planet:

$M=\frac{FR^2}{Gm}=\frac{(1250)(2.54\cdot 10^7)^2}{(6.67\cdot 10^{-11})(80.0)}=1.51\cdot 10^{26}kg$

Learn more about gravitational force:

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2 years ago

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Answer:

$y_{cg} = 1.03 m$

Explanation:

Given data:

weigh (head+arms + head) w_1 = 438 N

centre of gravity y_1= 1.28 m

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Center of gravity y_2 = 0.760 m

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4 years ago

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3 years ago

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SVEN [57.7K]

Answer:

The radius of the circle made by the person on the merry go round is 74.55 meters

Explanation:

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The radius of the circle made by the person on the merry go round = r

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$F_c = \dfrac{m \cdot v^2}{r} = m \cdot \omega ^2 \cdot r$

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