What is the gravitational force acting on a 70.0 kg object standing on the earth's surface ?

Answers

Answer 1

Answer: g( gravitational acceleration)=-9.81 m/s^2
m(mass) = 70.0 kg

According to Newton's second law of motion:

"F=ma"

Force is equal mass per acceleration therefore

W=mg
This is because weigth is the force of gravity acting in a body and "a" is replaced by "g" because "g"
is gravitational acceleration

Therefore:

W= mg
W= (70.0 kg) (-9.81m/s^2)
W= -686.7 N

* The "N" means newton which is (kg* m/s^2)

Answer 2

Answer:

Final answer:

The gravitational force acting on a 70.0 kg object on Earth is calculated using the formula w=mg and is equal to 686 N.

Explanation:

The gravitational force acting on a 70.0 kg object standing on the earth's surface can be calculated using the formula for weight (which is essentially the gravitational force on the object): w = mg, where m is the mass of the object and g is the acceleration due to gravity. On Earth, g is approximately 9.8 m/s².

Therefore, the gravitational force (weight) acting on this object can be calculated as follows: w = mg = (70.0 kg)(9.8 m/s²) = 686 N. Thus, the gravitational force on this 70.0 kg object standing on the earth's surface is 686 Newtons. It's important to understand that this force will vary if the object is moved to a location where g is different, like on the moon.

Learn more about Gravitational Force here:

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Answers

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How much force is needed to stop a 90-kg soccer player if he decelerates at 15 m/s^2?A. 15 N
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Answers

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Hope this helps!

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Answers

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D. a vacuum.

Answers

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Answers

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Answers

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