Answer:

Explanation:
t = Time taken = 
i = Current = 3 A
q(0) = Initial charge =
Charge is given by

The magnitude of the net electric charge of the capacitor is 
Answer:
The magnitude of magnetic field at given point =
×
T
Explanation:
Given :
Current passing through both wires = 5.0 A
Separation between both wires = 8.0 cm
We have to find magnetic field at a point which is 5 cm from any of wires.
From biot savert law,
We know the magnetic field due to long parallel wires.
⇒ 
Where
magnetic field due to long wires,
,
perpendicular distance from wire to given point
From any one wire
5 cm,
3 cm
so we write,
∴ 

![B =\frac{ 4\pi \times10^{-7} \times5}{2\pi } [\frac{1}{0.03} + \frac{1}{0.05} ]](https://tex.z-dn.net/?f=B%20%3D%5Cfrac%7B%204%5Cpi%20%5Ctimes10%5E%7B-7%7D%20%5Ctimes5%7D%7B2%5Cpi%20%7D%20%5B%5Cfrac%7B1%7D%7B0.03%7D%20%2B%20%5Cfrac%7B1%7D%7B0.05%7D%20%5D)

Therefore, the magnitude of magnetic field at given point = 
u= 215 km/hr = 215 * 1000/ 3600 = aprx 60m/s
v=0
t=2.7sec
v= u - at
u= at
60/2.7 = 22.23 m/s^2
Hope it helps
Answer:
At 81. 52 Deg C its resistance will be 0.31 Ω.
Explanation:
The resistance of wire =
Where
=Resistance of wire at Temperature T
= Resistivity at temperature T ![=\rho_0 \ [1 \ + \alpha\ (T-T_0\ )]](https://tex.z-dn.net/?f=%3D%5Crho_0%20%5C%20%5B1%20%5C%20%2B%20%5Calpha%5C%20%28T-T_0%5C%20%29%5D)
Where 
l=Length of the wire
& A = Area of cross section of wire
For long and thin wire the resistance & resistivity relation will be as follows

![\frac{0.25}{0.31}=\frac{1}{[1+\alpha(T-20)]}](https://tex.z-dn.net/?f=%5Cfrac%7B0.25%7D%7B0.31%7D%3D%5Cfrac%7B1%7D%7B%5B1%2B%5Calpha%28T-20%29%5D%7D)



T = 81.52 Deg C