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lukranit [14]
3 years ago
11

An object of weight W is in freefall close to the surface of Earth. The magnitude of the force that the object exerts on Earth i

s _____.
(A) greater than W
(B) less than W
(C) equal to W
(D) zero
(E) cannot be determined without knowing the relative masses of the object and the earth
Physics
2 answers:
Hoochie [10]3 years ago
4 0
Earth is B)equal to W 
atroni [7]3 years ago
4 0

Answer:

Equal to W

Explanation:

It is given that, an object of weight W is in free fall close to the surface of Earth. Weight of an object is equal to the product of its mass and acceleration. On the surface of Earth, the weight is given by :

W=mg

g=9.8\ m/s^2

The magnitude of the force that the object exerts on Earth is equal to W as per the Newton's third law of motion. The force exerted from one object to another object must be equal to and opposite to the force exerted from object 2 to 1.

So, the correct option is (c) "equal to W".

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Four point charges of magnitudes +3q, -q, +2q, and -4q are arranged in the corners of a square of side length L. The charge -q c
mafiozo [28]

Answer:

d) 0 V

Explanation:

It can be showed that the potential due to a point charge q, to a distance d from the charge, can be expressed as follows:

V = \frac{k*q}{r}

where k = \frac{1}{4*\pi*\epsilon0} = 9e9 N*m2/C2

As the potential is an scalar, and is linear with the charge, we can apply the superposition principle, which means that we can find the potential due to one of the charges, as if the other were not present.

By symmetry, all four charges are at the same distance from the center, so we can write the total potential, as follows:

V = \frac{k}{d} ( q1 + q2 + q3 + q4) (1)

where d, is the semi-diagonal of the square, that we can find applying Pythagorean theorem, as follows:

d = \sqrt{\frac{L^{2}}{4} + \frac{L^{2}}{4} } = L*\frac{\sqrt{2}}{2}

Replacing by the values in (1) we have:

V = \frac{9e9N*m2/C2}{\frac{L}{2}*\sqrt{2} }* ( +3q -q + 2q + -4q)  = 0 V

which is equal to the option d).

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