Electric field between the plates of parallel plate capacitor is given as

here area of plates of capacitor is given as


also the maximum field strength is given as

now we will plug in all data to find the maximum possible charge on capacitor plates


so the maximum charge that plate will hold will be given by above
Answer:
a. 
b. 
Explanation:
The inertia can be find using
a.





now to find the torsion constant can use knowing the period of the balance
b.
T=0.5 s

Solve to K'


Answer:
I think the answer is A
Explanation:
I need this brainliest answer please
Answer:
<em>Angular displacement=68.25 rad</em>
Explanation:
<u>Circular Motion</u>
If the angular speed varies from ωo to ωf in a time t, then the angular acceleration is given by:

The angular displacement is given by:

The wheel decelerates from ωo=13.5 rad/s to ωf=6 rad/s in t=7 s, thus:



Thus, the angular displacement is:



Angular displacement=68.25 rad