Two water balloons of mass 0.75 kg collide and bounce off of each other without breaking. Before the collision, one water balloo
1 answer:
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Answer:
change in entropy is 1.44 kJ/ K
Explanation:
from steam tables
At 150 kPa
specific volume
Vf = 0.001053 m^3/kg
vg = 1.1594 m^3/kg
specific entropy values are
Sf = 1.4337 kJ/kg K
Sfg = 5.789 kJ/kg
initial specific volume is calculated as
FROM STEAM Table
at 200 kPa
specific volume
Vf = 0.001061 m^3/kg
vg = 0.88578 m^3/kg
specific entropy values are
Sf = 1.5302 kJ/kg K
Sfg = 5.5698 kJ/kg
constant volume so
Change in entropy
=3( 3.36035 - 2.88) = 1.44 kJ/kg
Answer:
a) The magnitude of the force is 968 N
b) For a constant speed of 30 m/s, the magnitude of the force is 1,037 N
Explanation:
<em>NOTE: The question b) will be changed in other to give a meaningful answer, because it is the same speed as the original (the gallons would be 1.9, as in the original).</em>
Information given:
d = 106 km = 106,000 m
v1 = 28 m/s
G = 1.9 gal
η = 0.3
Eff = 1.2 x 10^8 J/gal
a) We can express the energy used as the work done. This work has the following expression:
Then, we can derive the magnitude of the force as:
b) We will calculate the force for a speed of 30 m/s.
If the force is proportional to the speed, we have: