The magnitude of the force required to stop the weight in 0.333 seconds is 67.6 N.
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Magnitude of required force to stop the weight</h3>
The magnitude of the force required to stop the weight in 0.333 seconds is calculated by applying Newton's second law of motion as shown below;
F = ma
F = m(v/t)
F = (mv)/t
F = (5 x 4.5)/0.333
F = 67.6 N
Thus, the magnitude of the force required to stop the weight in 0.333 seconds is 67.6 N.
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Answer:
Torque, 
Explanation:
It is given that,
Length of the wrench, l = 0.5 m
Force acting on the wrench, F = 80 N
The force is acting upward at an angle of 60.0° with respect to a line from the bolt through the end of the wrench. We need to find the torque is applied to the nut. We know that torque acting on an object is equal to the cross product of force and distance. It is given by :
So, the torque is applied to the nut is 34.6 N.m. Hence, this is the required solution.
If we consider Boyles law for gases, it states the following equation,
PV/T =k
where k is a constant
P-pressure, T- temperature and V -volume
the volume is constant at both situations as its a rigid tank as mentioned in the questions.
Therefore we consider Volume to be constant , then equation is
P/T = kV
kV = c (new constant)
P/T = c
P = cT
Therefore pressure is inversely proportional to temperature, whatever change in pressure would cause the same kind of change in temperature as well.
therefore when T decreases, P also decreases.
Answer is B
