both parts please. Will rate 5 An 9.00-kg point mass and a 14.0-kg point mass ar

Question

both parts please. Will rate 5

An 9.00-kg point mass and a 14.0-kg point mass are held in place 50.0 cm apart. A particle of mass m is released from a point between the two masses 15.0 cm from the 9.00-kg mass along the line connecting the two fixed masses. Part A Find the magnitude of the initial acceleration of the particle. Express your answer with the appropriate units. Part B What is the direction of the initial acceleration of the particle? smallCircle Acceleration of the particle is toward the 14.0-kg mass. smallCircle Acceleration of the particle is toward the 9.00-kg mass.

Explanation / Answer

to find acceleration you find the the force acting on the point where the third point mass is dropped

F = ma

then a = F/m

F = G mm/r^2
G = 6.673 * 10^-11

change cm to m

50 cm = 0.500 m
15.0 cm = 0.150 m

F9 = G * 9*m/0.15^2

F13 = G * 14*m/35^2

Fnet = G*m *(9/0.15^2 - 14/0.35^2)

Fnet = G*m*(9*0.35^2-14*0.15^2/(0.15^2*0.35^2).

a = Fnet/m = G*(9*0.35^2 - 14*0.15^2/(0.15^2*0.35^2))
a = G* 40.816

a = 2.72 * 10 ^-8 m/s^2

Part B

towards the 14 kg

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