Explanation:
The given data is as follows.
C = 
R = 
ohm
C
Q = 
Formula to calculate the time is as follows.
![Q_{t} = Q_{o} [e^{\frac{-t}{\tau}]](https://tex.z-dn.net/?f=Q_%7Bt%7D%20%20%3D%20Q_%7Bo%7D%20%5Be%5E%7B%5Cfrac%7B-t%7D%7B%5Ctau%7D%5D)
![13.5 \times 10^{-6} = 100 \times 10^{-6} [e^{\frac{-t}{2}}]](https://tex.z-dn.net/?f=13.5%20%5Ctimes%2010%5E%7B-6%7D%20%3D%20100%20%5Ctimes%2010%5E%7B-6%7D%20%5Be%5E%7B%5Cfrac%7B-t%7D%7B2%7D%7D%5D)
0.135 = 
= 7.407
t = 4.00 s
Therefore, we can conclude that time after the resistor is connected will the capacitor is 4.0 sec.
Answer:
C
Explanation:
It says is it a good idea the person so 1 person can say no and the other one can say yes so it is asks for a opinion
Answer:
10.1 N
Explanation:
Your answer is 10.1 N, I don't actually know how to do it but I hope it helps.
Answer:
d = V/E
Explanation:
From the definition, we can say that the electric field strength between the plates of a parallel plate capacitor is
E = v/d
where
E = electric field strength
V = potential difference
d = distance between the plates
On rearranging the equation and making d subject of the formula, we have
d = V/E
From the question, we're given that
V = 112 V
E = 1.12 kV/cm converting to V/m, we have 110000 V/cm
d = 112 / 110000
d = 0.00102 m
d = 1.02*10^-3 m
