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AfilCa [17]
3 years ago
7

Two coils are placed close together in a physics lab to demonstrate Faraday’s law of induction. A current of 5.00 A in one is sw

itched off in 1.00 ms, inducing a 9.00 V emf in the other. What is their mutual inductance?
Physics
1 answer:
Mariana [72]3 years ago
7 0

Answer:

Mutual inductance will be M=1.8\times 10^{-3}Hennry

Explanation:

We have given induced current \Delta i=5A

Time is given as t=1ms=10^{-3}sec

We have to find the mutual inductance between the coils

Induced emf is given as e = 9 volt

We know that induced emf is given by

e=M\frac{\Delta i}{\Delta t}

9=M\times \frac{5}{10^{-3}}

M=1.8\times 10^{-3}Hennry

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You are an astronaut floating alongside your space ship. If you throw a baseball with a speed of 10 m/s, you will move at:
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D

Explanation:

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13. Now we will examine the electric field of a dipole. The magnitude and direction of the electric field depends on the distanc
Gennadij [26K]

This question is incomplete, the complete question is;

Now we will examine the electric field of a dipole. The magnitude and direction of the electric field depends on the distance and the direction. We will investigate in detail just two directions. With charges available in the simulation (all the charges are either positive or negative 1 nC increments).

how do you create a dipole with dipole moment 1 x 10-9 Cm with a direction for the dipole moment pointing to the right. Make a table below that shows the amounts of  charge and the distance between the charges. There are many correct answers

Answer:

Given the data in question;

Dipole moment P = 1 × 10⁻⁹ C.m

now dipole pointing to the right;

               P→

_{-\theta } (-) ---------------->(+) _{+\theta }

               d

so let distance between the dipoles be d

∴ P = d\Theta

Let \Theta_{1} = 1 nC

so

P = d\Theta

1 × 10⁻⁹ =  1 × 10⁻⁹  × d

d = (1 × 10⁻⁹) / (1 × 10⁻⁹)

d = 1 m

Also Let \Theta_{2} = 2 nC

so

P = d\Theta

1 × 10⁻⁹ =  2 × 10⁻⁹  × d

d = (1 × 10⁻⁹) / (2 × 10⁻⁹)  

d = 0.5 m

Also Let \Theta_{3} = 3 nC

so

P = d\Theta

1 × 10⁻⁹ =  3 × 10⁻⁹  × d

d = (1 × 10⁻⁹) / (3 × 10⁻⁹)

d = 0.33 m

such that;

charge                 distance

1 nC                        1.00 m      

2 nC                       0.50 m

3 nc                        0.33 m

4 nC                       0.25 m

5 nC                       0.20 m      

5 0
3 years ago
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