Answer:
The work flow required by the compressor = 100.67Kj/kg
Explanation:
The solution to this question is obtained from the energy balance where the initial and final specific internal energies and enthalpies are taken from A-17 table from the given temperatures using interpolation .
The work flow can be determined using the equation:
M1h1 + W = Mh2
U1 + P1alph1 + ◇U + Workflow = U2 + P2alpha2
Workflow = P2alpha2 - P1alpha1
Workflow = (h2 -U2) - (h1 - U1)
Workflow = ( 684.344 - 491.153) - ( 322.483 - 229.964)
Workflow = ( 193.191 - 92.519)Kj/kg
Workflow = 100.672Kj/kg
After a thorough research, there exists the same question that has choices and the link of the graph (http://i37.servimg.com/u/f37/16/73/53/52/graph410.png)
<span>Choices:
A. 160 meters
B. 80 meters
C. 40 meters
D. 20 meters
E. 0 meters
</span>
The correct answer is letter E. 0 meters.
Explanation:
Weak nuclear force:
The interaction between the subatomic particles is called weak nuclear force.
The weak nuclear force is one of the four fundamental forces.
The weak nuclear force is effective at very short distance.
The range and relative strength of weak nuclear force is 10⁻¹⁸ m and 10²⁵ with respect to gravitational force respectively
Deuterium is formed due to the fusion of protons and neutrons under the action the weak force.
Example : Beta decay
Electromagnetic force:
The interaction between the charged particles is called electromagnetic force.
The electromagnetic force is one of the four fundamental forces.
The electromagnetic force is effective at long range distance.
The range and relative strength of electromagnetic force is infinity and 10³⁶ with respect to gravitational force respectively
Example : light
Answer:
Along path BC of the Otto cycle, heat transfer Qh into the gas occurs at constant volume, causing a further increase in pressure and temperature. This process corresponds to burning fuel in an internal combustion engine, and takes place so rapidly that the volume is nearly constant.
Answer:2.55 rad/s
Explanation:
Given
Diameter of ride=5 m
radius(r)=2.5 m
Static friction coefficient range=0.60-1
Here Frictional force will balance weight
And limiting frictional force is provided by Centripetal force

weight of object=mg
Equating two
f=mg



