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bearhunter [10]
4 years ago
15

An induced emf is produced inA. a closed loop of wire when it remains at rest in a nonuniform static magnetic field.B. a closed

loop of wire when it remains at rest in a uniform static magnetic field.C. a closed loop of wire moving at constant velocity in a nonuniform static magnetic field.D. only b and c above.E. all of the above.
Physics
1 answer:
Kazeer [188]4 years ago
7 0

Answer:

C. a closed loop of wire moving at constant velocity in a nonuniform static magnetic field.

Explanation:

As we know by Faraday's law that rate of change in magnetic flux will induce EMF

so mathematically we can say that

EMF = \frac{d\phi}{dt}

so we will have

\phi = B.A

so EMF will be induced if the flux linked with the closed loop will change with time.

So here correct answer will be

C. a closed loop of wire moving at constant velocity in a nonuniform static magnetic field.

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Answer:

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2 years ago
An archer defending a castle is on a 15.5 m high wall. He shoots an arrow straight down at 22.8 m/s. How much time does it take
vazorg [7]

Answer: 1.907

Explanation:

I did the math

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4 years ago
Hawks and gannets soar above the ground and, when they spot prey, they fold their wings and essentially drop like a stone. They
denis-greek [22]

Answer:

  v = 54.2 m / s

Explanation:

Let's use energy conservation for this problem.

Starting point Higher

         Em₀ = U = m g h

Final point. Lower

        Em_{f} = K = ½ m v²

        Em₀ = Em_{f}

        m g h = ½ m v²

         v² = 2gh

         v = √ 2gh

Let's calculate

         v = √ (2 9.8 150)

         v = 54.2 m / s

3 0
3 years ago
Where is oceanic crust being formed?
Sever21 [200]
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#CarryOnLearning

Butbot na dili na tinuod ga timala ra ko ana
3 0
3 years ago
Two resistors, of R1 = 3.11 Ω and R2 = 6.15 Ω, are connected in series to a battery with an EMF of 24.0 V and negligible interna
Firlakuza [10]
Part (a):
1- Since the resistors are in series, therefore, the total resistance is the summation of the two resistors.
Therefore:
Rtotal = R1 + R2 = 3.11 + 6.15 = 9.26 ohm

2- Since the two resistors are in series, therefore, the current flowing in both is the same. We will use ohm's law to get the current as follows:
V = I*R
V is the voltage of the battery = 24 v
I is the current we want to get
R is the total resistance = 9.26 ohm
Therefore:
24 = 9.26*I
I = 24 / 9.26
I = 2.59 ampere

Part (b):
To get the voltage across the second resistor, we will again use Ohm's law as follows:
V = I*R
V is the voltage we want to get
I is the current in the second resistor = 2.59 ampere
R is the value of the second resistor = 6.15 ohm
Therefore:
V = I*R
V = 2.59 * 6.15
V = 15.9285 volts

Hope this helps :)
7 0
3 years ago
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