Can somebody help please ! a. -8.3 m/s b.-4.2 m/s c.-0.12 m/s d. 0 m/s

If the distance between two masses is tripled, the gravitational force between changes by a factor of:_______

adell [148]

Answer:

option A

Explanation:

given,

distance between two masses is doubled

new distance, r' = 3 r

using gravitational force equation

$F = \dfrac{GMm}{r^2}$ ............(1)

new gravitational force

$F' = \dfrac{GMm}{r'^2}$

now from the given condition

$F' = \dfrac{GMm}{(3r)^2}$

$F' = \dfrac{GMm}{9r^2}$

$F' = \dfrac{1}{9}\dfrac{GMm}{r^2}$

now, from equation (1)

$F' = \dfrac{1}{9}F$

$\dfrac{F'}{F} = \dfrac{1}{9}$

now, the change in gravitational force factor is equal to $\dfrac{1}{9}$

Hence, the correct answer is option A

3 0

3 years ago

Which of the following BEST explains how good body composition is attained?

Nonamiya [84]

Answer:

C. balancing the amount of energy that is taken in with the amount of energy that is released by the body

Explanation:

im smart

6 0

4 years ago

What are two ways in which all types of precipitation are alike

Digiron [165]

1. they all come from the clouds 2.they are all formed by precipatation. hope this helps enjoyyyyyy!!!!!!!!!!!!!

7 0

3 years ago

The tension in the rope is constant and equal to 40 N as the block is pulled. What is the instantaneous power (in W) supplied by

Eduardwww [97]

Complete Question:

A 10 kg block is pulled across a horizontal surface by a rope that is oriented at 60° relative to the horizontal surface.

The tension in the rope is constant and equal to 40 N as the block is pulled. What is the instantaneous power (in W) supplied by the tension in the rope if the block when the block is 5 m away from its starting point? The coefficient of kinetic friction between the block and the floor is 0.2 and you may assume that the block starting at rest.

Answer:

Power = 54.07 W

Explanation:

Mass of the block = 10 kg

Angle made with the horizontal, θ = 60°

Distance covered, d = 5 m

Tension in the rope, T = 40 N

Coefficient of kinetic friction, $\mu = 0.2$

Let the Normal reaction = N

The weight of the block acting downwards = mg

The vertical resolution of the 40 N force, $f_{y} = 40sin \theta$

$\sum f(y) = 0$

$N + 40 sin \theta - mg = 0\\N = -40sin60 + 10*9.81 = 0\\N = 63.46 N$

$\sum f(x) = 0$

$40 cos 60 - f_{r} - ma = 0\\ f_{r} = \mu N\\ f_{r} = 0.2 * 63.46\\ f_{r} = 12.69 N\\40cos 60 - 12.69-10a = 0\\7.31 = 10a\\a = 0.731 m/s^{2}$

$v^{2} = u^{2} + 2as\\u = 0 m/s\\v^{2} = 2 * 0.731 * 5\\v^{2} = 7.31\\v = \sqrt{7.31} \\v = 2.704 m/s$

Power, $P = Fvcos \theta$

$P = 40 *2.704 cos60\\P = 54.074 W$

7 0

4 years ago

At ground level g is 9.8m/s^2. Suppose the earth started to increase its angular velocity. How long would a day be when people o

velikii [3]

Let s = rate of rotation
Let r = radius of earth = 6,400km 
Then solving (s^2) r = g will give the desired rate, from which length of day is inferred. 
People would not be thrown off. They would simply move eastward in a straight line while the curved surface of earth fell away from beneath them.

5 0

4 years ago

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