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Elina [12.6K]
3 years ago
11

Three identical very dense masses of 3500 kg each are placed on the x axis. One mass is at x1 = -100 cm , one is at the origin,

and one is at x2 = 320 cm . Part A What is the magnitude of the net gravitational force Fgrav on the mass at the origin due to the other two masses? Take the gravitational constant to be G = 6.67×10−11 N⋅m2/kg2 . Express your answer in newtons to three significant figures. Part B What is the direction of the net gravitational force on the mass at the origin due to the other two masses? +x direction or -x direction.
Physics
1 answer:
larisa86 [58]3 years ago
7 0

Answer:

<em>A) 7.37 x 10^-4 N</em>

<em>B) The resultant force will be towards the -x axis</em>

Explanation:

The three masses have mass = 3500 kg

For the force of attraction between the mass at the origin and the mass -100 cm away:

distance r = 100 cm = 1 m

gravitational constant G= 6.67×10^−11 N⋅m^2/kg^2

Gravitational force of attraction F_{g} = \frac{Gm^{2} }{r^{2} }

where G is the gravitational constant

m is the mass of each of the masses

r is the distance apart = 1 m

substituting, we have

F_{g} = \frac{6.67*10^{-11}*3500^{2} }{1^{2} } = 8.17 x 10^-4 N

For the force of attraction between the mass at the origin and the mass 320 cm away

distance r = 320 cm = 3.2 m

F_{g} = \frac{Gm^{2} }{r^{2} }

substituting, we have

F_{g} = \frac{6.67*10^{-11}*3500^{2} }{3.2^{2} } = 7.98 x 10^-5 N

Resultant force = (8.17 x 10^-4 N) - (7.98 x 10^-5 N) = <em>7.37 x 10^-4 N</em>

<em></em>

<em>B) The resultant force will be towards the -x axis</em>

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v_{o} - Initial velocity of the softball, measured in meters per second.

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The initial velocity of the softball is 14.711 meters per second.

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