A 5 kg sledge hammer is held over head (2 meters above the ground). What is the maximum amount of work it can do if it falls to
1 answer:
The maximum amount of work that the hammer can do while falling to the ground is equal to its variation of gravitational potential energy, which is given by:
where m is the mass of the hammer, g is the gravitational acceleration, h is the variation in altitude of the hammer.
Plugging the data of the problem inside the equation, we find
where m is the mass of the hammer, g is the gravitational acceleration, h is the variation in altitude of the hammer.
Plugging the data of the problem inside the equation, we find
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Answer:
The expression is Vmem = ΔV - I*Rax
Explanation:
According to the picture, if switch S is closed, the Cmem will be short. If Vcap = 0, the current flows through the capacitor only (not through Rmem), thus Vmem = 0 after closing S.
When C is fully charged, then we have:
Vmem = ΔV - I*Rax
Answer:
the tension in each side of the cable is 3677.57 N
Explanation:
given data
traffic light = 20 kg
cable between two poles = 30 m
sag in the cable = 0.40 m
solution
by the free body diagram
tan θ = .............1
θ = 1.527 °
and
tension = mg
The net force is along x - axis is express as
T2 cosθ = T1 cosθ .................2
so T2 - T1 ..............3
and
when we take it along y axis that is express as
( T1 + T2) sinθ = mg ...................4
so by equation 3 we put here
2 × T1 sin(1.527) = 20 × 9.8
T1 = 3677.57 N
