- Distance=2.5m
- Time=9.1s
Average Velocity=Total Distance/Total Time
A person exerts a horizontal force of F=45N on the end of an 86cm wide door. The magnitude of the torque due to F about the pivo
t point is determined by |τ|=|rxF|=rFsinθ . Determine the magnitude of the torque, |τ| , on the door about its hinges due to F . |τ|=0Nm |τ|=38.7Nm |τ|=3870Nm
1 answer:
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Answer:
The work energy added to the gas is
W = -72.48472 kJ
Explanation:
For constant pressure process
Work is given W = p×
However dH = m× = 1875 J
For mass m, dQ = 2625 J
That is m×1875 = 2625 or m = 2625/1875 = 1.4 kg
Therefore work doneper unit mass of gas is -W = p×(v₂ - v₁)
but pv = RT therefore work done per unit mass = -R×(T₂ - T₁) = -R×ΔT and
ΔT = 180.4K
Work done = -1.4×0.287×180.4 = -72.48472 kJ