Answer:
The amount of energy transferred to the water is 4.214 J
Explanation:
The given parameters are;
The mass of the object that drops = 5 kg
The height from which it drops = 86 mm (0.086 m)
The potential energy P.E. is given by the following formula
P.E = m·g·h
Where;
m = The mass of the object = 5 kg
g = The acceleration de to gravity = 9.8 m/s²
h = The height from which the object is dropped = 0.086 m
Therefore;
P.E. = 5 kg × 9.8 m/s² × 0.086 m = 4.214 J
Given that the potential energy is converted into heat energy, that raises the 1 g of water by 1°C, we have;
The amount of energy transferred to the water = The potential energy, P.E. = 4.214 J.
Answer:
Explanation:
Given
Airline flying at 34,000 ft.
Cabin pressurized to an altitude 8,000 ft.
We know that at standard condition ,density of air
We know that pressure difference
ΔP=ρ g ΔZ
Here ΔZ=34,000-8,000 ft
ΔZ=26,000 ft
ΔP=0.074 x 32.2 x 26,000
So pressure difference will be .
Answer:
(a) the rate of heat transfer to the coolant is Q = 139.71W
(b) the surface temperature of the shaft T = 40.97°C
(c) the mechanical power wasted by the viscous dissipation in oil 22.2kW
Explanation:
See explanation in the attached files
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