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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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Here is a force diagram of an object in water. The weight of the object is 15N and the buoyancy force is 17N. Will the object fl
lawyer [7]

Answer:

Object will float.

Explanation:

Total force on the body = Weight of body + Buoyancy force on body.

 Weight of body = 15 N downwards = 15 N

 Buoyancy force on body = 17 N upwards = -17 N

 Total force on body = 15 - 17 = -2 N = 2 N upwards

 So, the body will float.

Object will float.

8 0
3 years ago
What building materials do you believe would work well to build a home in that area? Explain why you chose these materials.
Luden [163]
Depends on what the area is. If it’s a rural place, Wood is cheep & easy to build. If there’s a lot of corrosion, strong weather/hurricane, bricks.
8 0
4 years ago
In the two-slit experiment, monochromatic light of frequency 5.00 × 1014 Hz passes through a pair of slits separated by 2.20 × 1
asambeis [7]

Explanation:

It is given that,

Frequency of monochromatic light, f=5\times 10^{14}\ Hz

Separation between slits, d=2.2\times 10^{-5}\ m

(a) The condition for maxima is given by :

d\ sin\theta=n\lambda

For third maxima,

\theta=sin^{-1}(\dfrac{n\lambda}{d})

\theta=sin^{-1}(\dfrac{n\lambda}{d})

\theta=sin^{-1}(\dfrac{nc}{fd})  

\theta=sin^{-1}(\dfrac{3\times 3\times 10^8\ m/s}{5\times 10^{14}\ Hz\times 2.2\times 10^{-5}\ m})  

\theta=4.69^{\circ}

(b) For second dark fringe, n = 2

d\ sin\theta=(n+1/2)\lambda

\theta=sin^{-1}(\dfrac{5\lambda}{2d})

\theta=sin^{-1}(\dfrac{5c}{2df})

\theta=sin^{-1}(\dfrac{5\times 3\times 10^8}{2\times 2.2\times 10^{-5}\times 5\times 10^{14}})

\theta=3.90^{\circ}

Hence, this is the required solution.

8 0
3 years ago
A hungry hawk was preying on a lizard who was running northwards to get away from the low-flying hawk. If the lizard can run 8m
Anastaziya [24]

Answer:

2m/s²

Explanation:

velocity = displacement (distance in a specified direction /time

8 0
3 years ago
4. Assume a multiple level queue system with a variable time quantum per queue, and that the incoming job needs 50ms to run to c
Troyanec [42]

Answer:

Explanation:

For the completion of incoming job it will take 50ms

First queue takes 5ms quantum time and the subsequent queue takes double of the previous question

So,

First queue T_1 = 5ms

Second queue T_2 = 2 × T_1 = 2 × 5 = 10ms

Third queue T_3 = 2 × T_2 = 2 × 10 = 20ms

Fourth queue T_4 = 2 × T_3 = 2 × 20 = 40ms

Fifth queue T_5= 2 × T_4 = 2 × 40 = 80ms.

Now, the job will be done after the fifth queue.

So, after the first queue, the job is not completed, so, we have first interruption

After the second queue, the job is not completed, so, we have second interruption

After the third queue, the job is not completed, so we have third interruption.

After the fourth queue, the job is not yet completed, so we have the fourth interruption

And in the fifth queue the job is completed, so we don't have any interruption here.

So, the job will be interrupted 4 times and it will finished on the fifth queue.

6 0
4 years ago
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