Answer:
Explanation:
Given that,
The mutual inductance of the two coils is
M = 300mH = 300 × 10^-3 H
M = 0.3 H
Current increase in the coil from 2.8A to 10A
∆I = I_2 - I_1 = 10 - 2.8
∆I = 7.2 A
Within the time 300ms
t = 300ms = 300 × 10^-3
t = 0.3s
Second Coil resistance
R_2 = 0.4 ohms
We want to find the current in the second coil,
The same induced EMF is in both coils, so let find the EMF,
From faradays law
ε = Mdi/dt
ε = M•∆I / ∆t
ε = 0.3 × 7.2 / 0.3
ε = 7.2 Volts
Now, this is the voltage across both coils,
Applying ohms law to the second coil, V=IR
ε = I_2•R_2
0.72 = I_2 • 0.4
I_2 = 0.72 / 0.4
I_2 = 1.8 Amps
The current in the second coil is 1.8A
energy. It will not disrupt the picture developing process by overexposing too much light on the film.
Answer: 
Explanation:
Given
Magnitude of charge is 
Force experienced is 
Electric field intensity is the electrostatic force per unit charge

Thus, the electric field intensity is 
Answer:
0.358Kg
Explanation:
The potential energy in the spring at full compression = the initial kinetic energy of the bullet/block system
0.5Ke^2 = 0.5Mv^2
0.5(205)(0.35)^2 = 12.56 J = 0.5(M + 0.0115)v^2
Using conservation of momentum between the bullet and the block
0.0115(265) = (M + 0.0115)v
3.0475 = (M + 0.0115)v
v = 3.0475/(M + 0.0115)
plugging into Energy equation
12.56 = 0.5(M + 0.0115)(3.0475)^2/(M + 0.0115)^2
12.56 = 0.5 × 3.0475^2 / ( M + 0.0115 )
12.56 = 0.5 × 9.2872/ M + 0.0115
12.56 = 4.6436/ M + 0.0115
12.56 ( M + 0.0115 ) = 4.6436
12.56M + 0.1444 = 4.6436
12.56M = 4.6436 - 0.1444
12.56 M = 4.4992
M = 4.4992÷12.56
M = 0.358 Kg
The average speed would be 33.29m/s.The average speed equation is:

First you will need to solve for the distance you traveled in each scenario. So we can solve this by getting the product of speed and the time traveled.
Scenario 1:
Speed = 29m/s
Time = 120s
Distance = ?
Distance = (29m/s)(120s)
= 3,480m
Scenario 2
Speed = 35m/s
Time = 300s
Distance = ?
Distance = (35m/s)(300s)
= 10,500m
Now that you have the distance of both, you can solve for your average speed.