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:
connecting two independent clauses
Answer:
Mechanical Efficiency = 83.51%
Explanation:
Given Data:
Pressure difference = ΔP=1.2 Psi
Flow rate = 
Power of Pump = 3 hp
Required:
Mechanical Efficiency
Solution:
We will first bring the change the units of given data into SI units.

Now we will find the change in energy.
Since it is mentioned in the statement that change in elevation (potential energy) and change in velocity (Kinetic Energy) are negligible.
Thus change in energy is

As we know that Mass = Volume x density
substituting the value
Energy = Volume * density x ΔP / density
Change in energy = Volumetric flow x ΔP
Change in energy = 0.226 x 8.274 = 1.869 KW
Now mechanical efficiency = change in energy / work done by shaft
Efficiency = 1.869 / 2.238
Efficiency = 0.8351 = 83.51%
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Answer:
Explanation:
The detailed and careful step by step calculation and analysis is as shown with appropriate formula in the attached files