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oksian1 [2.3K]
3 years ago
12

The Sojourner Mars rover has a weight of 42.7 N on Mars where the acceleration due to gravity is 3.72 m/s2. What is Sojourner's

mass on mars? What is Sojourner's mass on Earth? What is Sojourner's weight on Earth?
Physics
1 answer:
yulyashka [42]3 years ago
3 0

1) 11.5 kg

2) 11.5 kg

3) 112.7 N

Explanation:

1)

The relationship between the mass of an object and its weight is given by the relationship:

W=mg

where:

W is the weight

m is the mass of the object

g is the acceleration due to gravity (the strength of the gravitational field at the location of the object)

In this problem, the object is on Mars.

Its weight is:

W = 42.7 N

While the acceleration due to gravity is:

g=3.72 m/s^2

Therefore, we can re-arrange the equation to find the mass of the rover:

m=\frac{W}{g}=\frac{42.7}{3.72}=11.5 kg

2)

Here we want to find the mass of the rover on Earth.

We have to identify the main difference between mass and weight:

- The mass of an object is an intrinsec property of the object, and it describes the "amount of matter" contained in the object

- The weight of an object is the force of gravity acting on the object itself

From the definitions above, it is clear that:

- The mass of an object is independent from its location, therefore it does not change if the object moves to another planet

- The weight of an object depends on its location, as it depends on the value of g, the acceleration due to gravity

Therefore, the mass of the rover on the Earth is the same as the mass of the rover on Mars:

m = 11.5 kg

3)

The weight of an object, as stated before, is the force of gravity acting on the object, and it is given by

W=mg

where

m is the mass of the object

g is the acceleration  due to gravity

On the surface of the Earth, the value of the acceleration due to gravity is

g=9.8 m/s^2

While the mass of the rover is

m = 11.5 kg

Therefore, the weight of the rover on Earth is

W=(11.5)(9.8)=112.7 N

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==================

This is what the arithmetic says IF the information in the question
is correct.

I don't know how true this is, and I certainly don't plan to test it,
but I have read that a current as small as  15 mA  through the
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If 15 mA can do it, and the sweaty electrician's resistance is
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