Objects with masses of 290 kg and a 590 kg are separated by 0.390 m. (o) Find th
ID: 1776854 • Letter: O
Question
Objects with masses of 290 kg and a 590 kg are separated by 0.390 m. (o) Find the net gravitational force exerted by these objects on a 56.0-kg object placed midway between them magnitude direction Select 294823 offers significantly from the correct answer. Rework your solution from the beginning and check each step carefuly, N (b) At what position (other than infinitely remote ones) can the 56.0-kg object be placed so as to experience a net force of m from the 590-kg mass Need Help? PesasehTe toa TeerExplanation / Answer
(a) Draw the digram -
(590 kg) ------0.195 m------ (56 kg) ------0.195 m------ (290 kg)
Now for the Left Half:
(590 kg) ------0.195 m------ (56 kg)
F = G M m / d^2 = (6.67x10^-11 x 590x56) / 0.195^2 = 5.80 x 10^-5 N
Now for the Right Half:
(56 kg) ------0.195 m------ (290 kg)
F = G N m / d^2 = (6.67x10^-11 x 290x56) / 0.195^2 = 2.85 x 10^-5 N
Therefore, the net force = 5.80 x 10^-5 N - 2.85 x 10^-5 N = 2.95 x 10^-5 N
(b) Again for this situation there can be an infinite solutions.
Means there is NOT one point in space where the net force will be zero.
Let us see that.
Here we will need to place one mass at any distance from the 56 kg mass. Then solve for the distance required to balance the forces. Otherwise the solution will cancel out your distances, "d's."
Assign a distance for the 590 kg mass as a variable "k". Thus the force will be
F = G ( 590 ) ( 56 ) / k^2
Assign a distance for the 290 kg mass as a variable "d". Thus the force will be
f = G (290) ( 56 ) / d^2
Solve
F = f
33040 / k^2 = 16240 / d^2, Rearranging
=> d^2*(33040 / k^2) = 16240
=> d = k*0.70
Now, setting the 590 kg mass any distance, k, from the 56 kg mass,
there will be the equivalent of placing the 290 kg mass a distance, d, away from the 56 kg mass,...
where d = k*0.70, in meters, and k can take the value of any number.
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