Physics, asked by naimabuaisha1056, 9 months ago

the mass of an object on moon is 60 kg how much is his mass on earth?

Answers

Answered by AwesomeSoul47
81

Answer:

\huge\mathfrak\red{Mass of an object}

Mass does not vary by the strength of the gravitational field. What changes is the weight which will only be about one-sixth of what it is on Earth.

So a 60Kg mass on the Moon would weigh approximately the same as a 10Kg mass on the Earth.

Explanation:

The mass of the man would remain the same

His weight would change. Assuming the man’s weight on earth is 60 N.

Since weight is a force, according to Newton’s second law,

F = ma. ( m = mass,)

( a = acceleration due to gravity)

First lets find the mass of the man,

as it is required to find his weight on the moon.

F = ma [taking a of earth as 10 m/s^2]

60 = m . 10 [divide 10 on both sides]

m = 60/10 = 6 Kg

Acceleration due to gravity on the moon is 83% less than the acceleration due to gravity on earth (1.622 m/s^2).

 F = ma

F = 6 . 1.622 = 9.732 N

So a person weighting 60 N on earth would approximately weight around 10 N on the moon.

Hope this helped! .

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Answered by raviprakashtiwari470
48

Answer:

\large\underline\green {\sf Mass \: of \: an \: object}

Mass does not vary by the strength of the gravitational field.

What changes is the weight which will only be about one-sixth of what it is on Earth.

So a 60Kg mass on the Moon would weigh approximately the same as a 10Kg mass on the Earth.

The mass of the man would remain the same…

His weight would change. Assuming the man’s weight on earth is 60 N.

Since weight is a force, according to Newton’s second law,

\underline\bold{ F \: = \: ma. \: m \:  = \: mass,}

( a = acceleration due to gravity)

First lets find the mass of the man,

as it is required to find his weight on the moon.

F = ma [taking a of earth as 10 m/s^2]

60 = m . 10 [divide 10 on both sides]

 m \:  = \:  60/10 \: =\: 6 \: Kg

Acceleration due to gravity on the moon is 83% less than the acceleration due to gravity on earth (1.622 m/s^2).

\underline\bold\pink{ F \: = \: ma }

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