Answer:
132 N
Explanation:
Given that a 1.1 kg hammer strikes a nail. Before the impact, the hammer is moving at 4.5 m/s; after the impact it is moving at 1.5 m/s in the opposite direction. If the hammer is in contact with the nail for 0.025 s, what is the magnitude of the average force exerted by the hammer on the nail
From Newton 2nd law of motion,
Change in momentum = impulse.
Change in momentum = m( V - U )
Substitute all the parameters into the formula
Change in momentum = 1.1 ( 4.5 - 1.5 )
Change in momentum = 1.1 × 3
Change in momentum = 3.3 kgm/s
Impulse = Ft
That is,
Ft = 3.3
Substitute time t into the formula above
F × 0.025 = 3.3
F = 3.3 / 0.025
F = 132 N
Therefore, the magnitude of the average force exerted by the hammer on the nail is 132 N.
A charged particle moving in a magnetic field experiences a force equal to:
Thus, the magnitude of the force that the proton experiences is given by:
The magnetic field is perpendicular to the proton's velocity, therefore, we have 
. Replacing the given values, we obtain:
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