A1.10-kg wrench is acting on a nut trying to turn it. the length of the wrench lies directly to the east of the nut. a force 150.0 n acts on the wrench at the position 15.0 cm from the center of the nut in a direction 30.0 degree north of east. what is the magnitude of the torque about the center of the nut ?

The problem can be solve using the formula

τ=r F sin(θ)

Where

τ is the torque,

r is the distance from where the pivot is to where the force is,

F is the force exerted, and

θ is the angle between the force and radius

τ=(0.15m) (150N) sin(30) = 11.25 Nm

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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