When Earth and the Moon are separated by adistance of 3.84 × 10^8 meters, the magnitude of
the gravitational force of attraction between
them is 2.0 × 10^20 newtons. What would be the
magnitude of this gravitational force of attraction
if Earth and the Moon were separated by a
distance of 1.92 × 10^8 meters?
(1) 5.0 × 10^19 N (3) 4.0 × 10^20 N
(2) 2.0 × 10^20 N (4) 8.0 × 10^20 N

Answers

Answer 1

Answer: Using the Universal Gratitation Law, we have:

$F= (MmG)/(d^2) \n MmG=2*10^(20)*(3.84*10^8)^2 \n MmG=29.4912*10^36$

Again applying the formula in the new situation, comes:

$F= (MmG)/(d^2) \n F= (29.4912*10^36)/((1.92*10^8)^2) \n \boxed {F=8*10^(20)}$

Number 4

If you notice any mistake in my english, please let me know, because i am not native.

Answer 2

Answer:
The strength of the gravitational forces between two masses is
inversely proportional to the square of the distance between them.

So if you change the distance to

   (1.92 x 10⁸) / (3.84 x 10⁸) = 1/2

of what it is now, then you would change the force to

   1 / (1/2)² = 4

of what it is now.

   (4) x (2 x 10²⁰) = 8.0 x 10²⁰ newtons .

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Answers

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A body oscillates with simple harmonic motion along the x-axis. Its displacement varies with time according to the equation x = 0.5 sin (pt+p/3). The acceleration (in m/s2) of the body at t = 1.0 s is approximatelya. 3.5
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Answers

I'll tell you how I look at this, although I may be missing something important.

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Speed = position' = x'(t) = 0.5 p cos(pt + p/3)

Acceleration = speed' = position ' ' = x ' '(t) = -0.5 p² sin(pt + p/3)

At (t = 1.0),

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Answers

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Answers

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