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
Answer:
13.23J
Explanation:
PE = m*g*h
PE = (3 kg ) * (9.8 m/s/s) * (0.45 m)
bonded pairs of electrons, lone pairs of electrons.
Answer:
1)
is<u> positive.</u>
<u></u>
2) 
Explanation:
<h2><u>
Part 1:</u></h2>
<u></u>
The charged rod is held above the balloon and the weight of the balloon acts in downwards direction. To balance the weight of the balloon, the force on the balloon due to the rod must be directed along the upwards direction, which is only possible when the rod exerts an attractive force on the balloon and the electrostatic force on the balloon due to the rod is attractive when the polarities of the charge on the two are different.
Thus, In order for this to occur, the polarity of charge on the rod must be positive, i.e.,
is <u>positive.</u>
<u></u>
<h2><u>
Part 2:</u></h2>
<u></u>
<u>Given:</u>
- Mass of the balloon, m = 0.00275 kg.
- Charge on the balloon,

- Distance between the rod and the balloon, d = 0.0640 m.
- Acceleration due to gravity,

In order to balloon to be float in air, the weight of the balloom must be balanced with the electrostatic force on the balloon due to rod.
Weight of the balloon, 
The magnitude of the electrostatic force on the balloon due to the rod is given by

is the Coulomb's constant.
For the elecric force and the weight to be balanced,

Answer:
16.9000000000000001 J
Explanation:
From the given information:
Let the initial kinetic energy from point A be
= 1.9000000000000001 J
and the final kinetic energy from point B be
= ???
The charge particle Q = 6 mC = 6 × 10⁻³ C
The change in the electric potential from point B to A;
i.e. V_B - V_A = -2.5 × 10³ V
According to the work-energy theorem:
-Q × ΔV = ΔK




