Answer:
1) The plane of the loop is perpendicular to the magnetic field.
2) The magnetic flux is independent of the orientation of the loop.p
Explanation:
The flux is calculated as φ=BAcosθ. The flux is therefore the highest when the magnetic field vector is perpendicular to the plane of the loop We can also deduce that the flux is zero when there is no magnetic field part perpendicular to the loop When the angle reaches zero, the flux is in the limit because when the angle becomes zero, the cos is the maximum.
Answer: scenario b and scenario c uses most power
Explanation:
Scenario a:
Work=120J
Time=8 seconds
Power=work ➗ time
Power=120 ➗ 8
Power=15
Power=15 watts
Scenario b:
Work=160J
Time=8 seconds
Power=work ➗ time
Power=160 ➗ 8
Power=20
Power =20 watts
Scenario c:
Work=200J
Time=10 seconds
Power= work ➗ time
Power=200 ➗ 10
Power=20
Power=20 watts
Scenario b and scenario c uses most power
Answer:
A. Final pressure P2
P2/P1 = (T2/T1)^n/n-1
P1 = 4bar
T1 = 438K
T2 = 300K
Polytropic index, n, = 1.3
P2 = 4 (300/438)^1.3/1.3-1
P2 = 4 (300/438)^4.333
P2 = 4 * 0.19400
P2 = 0.776bar.
B. The work done is;
W = mR/ n-1 (T1 -T2)
Where, R = 0.1889kJ/kg.K, m = 1
W = 1 * 0.1889/ 1.3-1 * (438-300)
W = 86.89kJ/kg.
C. The heat transfer, Q
Q = W + ΔU
Q = W + mCv(T2-T1), where Cv of nitrogen is 0.743kj/kgk
Q = 86.89 + 1 * 0.743 (300-438)
Q = 86.89 + (-102.534)
Q = -15.644kJ/K
Q = 15.64kJ/K
Answer:
Option C
Explanation:
In a large sized factory, it is essential to create cooling system based on duct work because it can then be able to regulate cooling of any section of the factory from one place. Also, ductwork cooling is preferred in large spaces such as big offices building, towers, factories etc.
Hence, option C is correct
