Answer:
Mostly yes
Explanation:
But you should consider,brand value, material used and lasting of the equipment also
HP stands for horsepower
It's mostly used in pumps
Answer:
Part a)

Part b)

Part c)
Since we know that the base area will remain same always
so here the length and width of the object is not necessary to obtain the above data in such type of questions
Explanation:
Part a)
As we know that when cylinder float in the water then weight of the cylinder is counter balanced by the buoyancy force
So here we know
buoyancy force is given as



Now we know that the weight of the cylinder is given as

now we have


Part b)
When the same cylinder is floating in other liquid then we will have

so we have


Part c)
Since we know that the base area will remain same always
so here the length and width of the object is not necessary to obtain the above data in such type of questions
Answer:
The the quality of the refrigerant at the exit of the expansion valve is 0.179.
Explanation:
Given that,
Initial pressure = 10 bar
Temperature = 22°C
Final pressure = 2.0 bar
We using the value of h

The refrigerant during expansion undergoes a throttling process
Therefore, 
We need to calculate the quality of the refrigerant at the exit of the expansion valve
At 2.0 bar,
The property of ammonia


Using formula

Put the value into the formula



Hence, The the quality of the refrigerant at the exit of the expansion valve is 0.179.