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goldfiish [28.3K]
3 years ago
9

Part b suppose the magnitude of the gravitational force between two spherical objects is 2000 n when they are 100 km apart. what

is the magnitude of the gravitational force fg between the objects if the distance between them is 150 km ? express your answer in newtons to three significant figures. hints fg = 889 n submitmy answersgive up correct significant figures feedback: your answer 890 n was either rounded differently or used a different number of significant figures than required for this part. part c what is the gravitational force fg between the two objects described in part b if the distance between them is only 50 km ?
Physics
1 answer:
guajiro [1.7K]3 years ago
4 0

The formula for gravitational force is:

F = G m1 m2 / r^2

where G m1 m2 are constants, therefore:

F r^2 = constant

 

Part b. Given F1 = 2000 N, r1 = 100 km

Find F2 = ?, r2 = 150 km

 

(2000 N) * (100 km)^2 = F2 * (150 km)^2

F2 = 888.89 N

 

 

Part c. Given F1 = 2000 N, r1 = 100 km

Find F2 = ?, r2 = 50 km

 

(2000 N) * (100 km)^2 = F2 * (50 km)^2

F2 = 8000 N

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A flat uniform circular disk (radius = 2.00 m, mass = 1.00 ✕ 102 kg) is initially stationary. The disk is free to rotate in the horizontal plane about a friction less axis perpendicular to the center of the disk. A 40.0-kg person, standing 1.25 m from the axis, begins to run on the disk in a circular path and has a tangential speed of 2.00 m/s relative to the ground.

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(b)t = 2.41 s

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0 = L for disk + L............... for runner

0 = Iω² - mv²r ...................they're opposite in direction

0 = (MR²/2)(ω²) - mv²r ................where is ω is angular speed which is required in part (a) of question

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b.)

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As the person and the disk are moving in opposite directions, the person will run part of a revolution and the turning disk would complete the whole revolution.

(angle) + (angle disk turns) = 2π

(1.6 rad/s)(t) + ωt = 2π

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t = 2.41 s

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