2 years ago

What change occurs to the gravitational force of attraction between two bodies when the

Physics

1 answer:

Lisa [10]2 years ago

4 0

Answer:

when the radius is halved, F becomes 4 times

Explanation:

The formula is given by

$F = \frac{Gm1m2}{r^2}$

When r is halved:

$F = \frac{Gm1m2}{(1/2r)^2}$

=> $F = \frac{Gm1m2}{1/4 r^2}$

=> $F = 4(\frac{Gm1m2}{r^2} )$

This means when the radius is halved, F becomes 4 times

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An object with velocity 141 ft/s has a kinetic energy of 1558.71 ftâˆ™lbf, on a planet whose gravity is 31.5 ft/s2. What is its

Sidana [21]

Answer:

The mass of the object is 5.045 lbm.

Explanation:

Given;

kinetic energy of the object, K.E = 1558.71 ft.lbf

velocity of the object, V = 141 ft/s

The kinetic energy of the object is calculated as;

$K.E = \frac{1}{2} mV^2\\\\mV^2 = 2K.E\\\\m = \frac{2K.E}{V^2} \\\\1 \ lbf = 32.174 \ lbm.ft/s^2\\\\m = \frac{2 \ \times \ 1558.71 \ ft.lbf \ \times \ 32.174 \ lbm.ft/s^2 }{(141 \ ft/s)2 \ \ \times \ \ \ \ 1 \ lbf\ }$

$m = \frac{(2 \ \times \ 1558.71 \ \times \ 32.174) \ lbm.ft^2/s^2 }{(141 )^2\ ft^2/s^2 }\\\\m = \frac{(2 \ \times \ 1558.71 \ \times \ 32.174) \ lbm }{(141 )^2 }\\\\m = 5.045 \ lbm$