Answer:
Isotope it will have a different number of neutrons than normal
Answer:
The power for circular shaft is 7.315 hp and tubular shaft is 6.667 hp
Explanation:
<u>Polar moment of Inertia</u>
= 0.14374 in 4
<u>Maximum sustainable torque on the solid circular shaft</u>
=
= 3658.836 lb.in
= lb.ft
= 304.9 lb.ft
<u>Maximum sustainable torque on the tubular shaft</u>
=
= 3334.8 lb.in
= lb.ft
= 277.9 lb.ft
<u>Maximum sustainable power in the solid circular shaft</u>
=
= 4023.061 lb. ft/s
= hp
= 7.315 hp
<u>Maximum sustainable power in the tubular shaft</u>
=
= 3666.804 lb.ft /s
= hp
= 6.667 hp
Answer:
Frequency required will be 2421.127 kHz
Explanation:
We have given inductance
Current in the inductor
Voltage v = 13 volt
Inductive reactance of the circuit
We know that
f = 2421.127 kHz
Based on the situation above the the work done was 400 Joules. <span>Q = FS cos(theta) is the so-called work function. It's important to learn the work physics; you'll see it over and over in science/physics class. Theta is the angle between the force vector F and the distance vector S. In your problem we assume theta = 0, the two vectors were assumed aligned.</span>