Since the stone was dropped from height, its initial velocity = 0 m/s
Using v² = u² + 2gs.
Where g ≈ 10 m/s², u = initial velocity = 0 m/s, s = height from drop = 2.5 m
v² = u² + 2gs
v² = 0² + 2*10*2.5
v² = 0 + 50
v² = 50
v = √50
v ≈ 7.07 m/s
Hence velocity just before hitting the ground is ≈ 7.07 m/s
Answer:
a. 572Btu/s
b.0.1483Btu/s.R
Explanation:
a.Assume a steady state operation, KE and PE are both neglected and fluids properties are constant.
From table A-3E, the specific heat of water is , and the steam properties as, A-4E:
Using the energy balance for the system:
Hence, the rate of heat transfer in the heat exchanger is 572Btu/s
b. Heat gained by the water is equal to the heat lost by the condensing steam.
-The rate of steam condensation is expressed as:
Entropy generation in the heat exchanger could be defined using the entropy balance on the system:
Hence,the rate of entropy generation in the heat exchanger. is 0.1483Btu/s.R
Answer:
it is very hard question for me sorry i cant solve it
D. Volume will increase
Volume and temperature are directly proportional for a gas.
Answer:
C. Increasing its buoyancy