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monitta
4 years ago
5

24. A 50-turn coil has a diameter of 15 cm. The coil is placed in a spatially uniform magnetic field of magnitude 0.50 T so that

the face of the coil and the magnetic field are perpendicular. Find the magnitude of the emf induced in the coil if the magnetic field is reduced to zero uniformly in (a) 0.10 s, (b) 1.0 s, and (c) 60 s.
Physics
1 answer:
finlep [7]4 years ago
4 0

Answer:

0.1 = 4.42v

1.0 =0.44v

60 = 7.4 × 10^-3 V

Explanation:

Apply the formula of:

Emf = (No of turns × magnetic flux)÷(the given time)

Where your magnetic flux = (0.50 × cross sectional area(pi × 0.075^2))

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Mrrafil [7]
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E=\frac{1}{2}kA^2
Where k is a constant that depends on the type of the wave you are looking at and A is amplitude.
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E_1=\frac{1}{2}k(1)^2=\frac{1}{2}k\\E_2=\frac{1}{2}k(2)^2=\frac{4}{2}k=2k\\
We can see that energy associated with the wave is 4 times smaller when we decrease its amplitude by half. So the answer should be C.


6 0
3 years ago
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A proton moves at constant velocity in the direction, through a region in which there is an electric field and a magnetic field.
dimaraw [331]

Answer:

F_{net}=0

Explanation:

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We know that the net force in the region of magnetic and electric field is given by Lorentz forces. But here, the proton moves with constant velocity. So, the net force acting on it is 0.

i.e.

F_{net}=0

Hence, this is the required solution.

7 0
3 years ago
What three factors control the path of a surface current
KATRIN_1 [288]

Answer:

Surface currents are controlled by three factors: global winds, the Coriolis effect, and continental deflections. surface create surface currents in the ocean. Different winds cause currents to flow in different directions. objects from a straight path due to the Earth's rotation.

Explanation:

3 0
3 years ago
Ksp for agbr is 5x10-13. what is the maximum concentration of silver ion that you can have in a 0.1 m solution of nabr?
liberstina [14]

Answer : The maximum concentration of silver ion is 5\times 10^{-12}m

Solution : Given,

K_{sp} for AgBr = 5\times 10^{-13}

Concentration of NaBr solution = 0.1 m

The equilibrium reaction for NaBr solution is,

NaBr(aq)\rightleftharpoons Na^++Br^-

The concentration of NaBr solution is 0.1 m that means,

[Na^+]=[Br^-]=0.1m

The equilibrium reaction for AgBr is,

                          AgBr\rightleftharpoons Ag^++Br^-

At equilibrium                     s       s

The expression for solubility product constant for AgBr is,

K_{sp}=[Ag^+][Br^-]

The concentration of Ag^+ = s

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Now put all the given values in K_{sp} expression, we get

5\times 10^{-13}=(s)(0.1+s)

By rearranging the terms, we get the value of 's'

s=5\times 10^{-12}m

Therefore, the maximum concentration of silver ion is 5\times 10^{-12}m.

4 0
3 years ago
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' C ' is the only correct statement on the list.  We don't know anything about diagram-x or diagram-y because we can't see them.

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