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Lyrx [107]
3 years ago
6

Two coils of wire are placed close together. Initially, a current of 1.80 A exists in one of the coils, but there is no current

in the other. The current is then switched off in a time of 4.02 x 10-2 s. During this time, the average emf induced in the other coil is 4.38 V. What is the mutual inductance of the two-coil system?

Physics
2 answers:
sattari [20]3 years ago
8 0

Answer: the mutual inductance is - 0.0978H

Explanation:

Detailed explanation and calculation is shown in the image below

STatiana [176]3 years ago
6 0

Answer:

M = 0.1H

Explanation:

Please see attachment below.

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It would be risk factor
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2 years ago
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A mountain climber, in the process of crossing between two cliffs by a rope, pauses to rest. She weighs 514 N. As the drawing sh
n200080 [17]

Answer:

Part a)

T_L = 155.4 N

Part b)

T_R = 379 N

Explanation:

As we know that mountain climber is at rest so net force on it must be zero

So we will have force balance in X direction

T_L cos65 = T_R cos80

T_L = 0.41 T_R

now we will have force balance in Y direction

mg = T_L sin65 + T_Rsin80

514 = 0.906T_L + 0.985T_R

Part a)

so from above equations we have

514 = 0.906T_L + 0.985(\frac{T_L}{0.41})

514 = 3.3 T_L

T_L = 155.4 N

Part b)

Now for tension in right string we will have

T_R = \frac{T_L}{0.41}

T_R = 379 N

3 0
3 years ago
A car goes from 5 m/s to 25 m/s in 6 s. What is the acceleration of the car?
damaskus [11]

Answer:

5m/s

Explanation:

3 0
3 years ago
Some scientists believe that the chain of underwater mountains beneath the Pacific, which stretches away from Hawaii up to Alask
solong [7]

Answer: Erosion or water level rising

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7 0
3 years ago
The rectangular plates in a parallel-plate capacitor are 0.063 m x 5.4 m. A distance of 3.5 x 10^-5m separate the plates. The pl
Aleks04 [339]

The capacitance of the capacitor is 1.8\cdot 10^{-9}F

Explanation:

The capacitance of a parallel-plate capacitor is given by the equation

C=\frac{k\epsilon_0 A}{d}

where

k is the dielectric constant of the medium

\epsilon_0 = 8.85\cdot 10^{-12} F/m is the vacuum permittivity

A is the area of the plates

d is the separation between the plates

For the capacitor in this problem, we have:

k = 2.1 is the dielectric constant

d=3.5\cdot 10^{-5}m is the separation between the plates

A=0.063m \cdot 0.054m = 0.0034 m^2 (I assumed that 5.4 m is a typo, since it is not a realistic size for the side of the plate)

Therefore, the capacitance of the  capacitor is

C=\frac{(2.1)(8.85\cdot 10^{-12})(0.0034)}{3.5\cdot 10^{-5}}=1.8\cdot 10^{-9}F

Learn more about capacitors:

brainly.com/question/10427437

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brainly.com/question/9617400

#LearnwithBrainly

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