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Nesterboy [21]
3 years ago
15

The magnetic flux through a coil of wire containing two loops changes at a constant rate from -83 Wb to 82 Wb in 0.39 s .

Physics
1 answer:
Serga [27]3 years ago
6 0

Answer:

423v

Explanation:

Using

E= -N ∆န/ ∆ t

= ( -1) X (-83 - 82)/0.39

= 423volts

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Levart [38]
The electric potential energy of the charge is equal to the potential at the location of the charge, V, times the charge, q:
U=qV
The potential is given by the magnitude of the electric field, E, times the distance, d:
V=Ed
So we have
U=qEd (1)
However, the electric field is equal to the electrical force F divided by the charge q:
E= \frac{F}{q}
Therefore (1) becomes
U=Fd
And if we use the data of the problem, we can calculate the electrical potential energy of the charge:
U=Fd=(3.6 \cdot 10^{-4}N)(9.8 \cdot 10^{-5} m)=3.53 \cdot 10^{-8} J
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A sinusoidal electromagnetic wave is propagating in vacuum. At a given point P and at a particular time, the electric field is i
olasank [31]

Answer:

a

The direction of the wave propagation is the negative  z -axis

b

The amplitude of  electric and magnetic field are  A_E= 3.35*10^5 V/m ,

A_M= 1.12 *10^{-3} T respectively

Explanation:

According to right hand rule, your finger (direction of electric field) would be pointing in the positive  x-axis  i.e towards your right let your palms be face toward the direction of the magnetic field i.e negative y-axis  (toward the ground ) Then anywhere your thumb stretched out is facing is the direction of propagation of the wave here in this case is the negative  z -axis

             The Intensity of the wave is mathematically represented as

                          I = \frac{1}{2} c \epsilon _O E_{rms}^2

Given that I = 7.43 \frac{kW}{cm^2} =  7.43 \frac{*10^3}{*10^-{4} }= 7.43*10^7 \frac{W}{m^2}

Making E_{rms} the subject we have

                   E_{rms} = \sqrt{\frac{I}{0.5*c*\epsilon_o} }

Substituting values as given on the question

                E_{rms} = \sqrt{\frac{7.43 *10^7[\frac{W}{m^2} ]}{0.5 * 3.08*10^8 *8.85*10^{-12}} }

                          = 2.37*10^5 \ V/m

The amplitude of the electric field is mathematically represented as

                  A_E = \sqrt{2} * E_{rms}

                         = \sqrt{2} * 2.37*10^5

                        A_E= 3.35*10^5 V/m

The amplitude of the magnetic field is mathematically represented as

                       A_M = \frac{A_E}{c}

Substituting value

                      A_M = \frac{3.35 *10^5}{3.0*10^8}

                             A_M= 1.12 *10^{-3} T

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A cat with a mass of 5 kg sits on a branch that is 26 m off the ground. What is
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Answer:

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Explanation:

P. E= mgh

where m = 5 kg

g = 9.8m/s²

h = 26m

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