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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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Calculate the work done (in J) by a 90.0 kg man who pushes a crate 4.25 m up along a ramp that makes an angle of 20.0° with the
Mrrafil [7]

Answer:

W = 3.4x0³ J.

Explanation:

The work done by the man is given by the following equation:

W = F_{t}\cdot d     (1)

<em>where W: is the work, Ft is the total force and d: is the displacement = 4.25 m.</em>  

We need to find first the total force Ft, which is:

Ft = Fm + W

<em>where Fm: is the force exerted by the man = 535 N, W: is the weight = m*g*sin(θ), m: is the mass of the man, g: is the gravitational acceleration = 9.81 m/s², and θ: is the angle = 20.0°.  </em>

F_{t} = Fm + W = 500 N + 90.0 kg*9.81 m/s^{2} * sin(20.0) = 802.0 N

Hence, the work is:

W = 802.0 N \cdot 4.25 m = 3.4 \cdot 10 ^{3} J  

Therefore, the work done by the man is 3.4x10³ J.  

I hope it helps you!      

8 0
3 years ago
Is aluminum foil reflecting onto something conduction, convection, or radiation?
mojhsa [17]
I had the SAME problem, put down Radiation and it’s thermal/light.
4 0
3 years ago
If you walk 5 km north and then 12 km east. Your resultant displacement is____.a. magnitude: 1.1km; direction: 53.1 degrees east
sweet [91]

Answer:

Resultant = 13km

Direction = 67.38° East of North

Explanation:

Given the following :

5km North ; 12km East

Resultant Displacement (r) :

r² = 5² + 12²

r² = 25 + 144

r² = 169

r = √169

r = 13

Direction:

Tangent = opposite / Adjacent

Tanθ = opposite / Adjacent

Opposite = 12 ; adjacent = 5

Tanθ = (12/5)

Tanθ = 2.4

θ = tan^-1(2.4)

θ = 67.38° east of north

4 0
3 years ago
A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is r
aivan3 [116]

Answer:

Explanation:

Given that,

Initial angular velocity is 0

ωo=0rad/s

It has angular velocity of 11rev/sec

ωi=11rev/sec

1rev=2πrad

Then, wi=11rev/sec ×2πrad

wi=22πrad/sec

And after 30 revolution

θ=30revolution

θ=30×2πrad

θ=60πrad

Final angular velocity is

ωf=18rev/sec

ωf=18×2πrad/sec

ωf=36πrad/sec

a. Angular acceleration(α)

Then, angular acceleration is given as

wf²=wi²+2αθ

(36π)²=(22π)²+2α×60π

(36π)²-(22π)²=120πα

Then, 120πα = 8014.119

α=8014.119/120π

α=21.26 rad/s²

Let. convert to revolution /sec²

α=21.26/2π

α=3.38rev/sec

b. Time Taken to complete 30revolution

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Then, t=120π/58π

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wf=wo+αt

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22π=21.26t

Then, t=22π/21.26

t=3.251seconds

d. Number of revolution to get to 11rev/s

∆θ= ½(wf+wo)•t

∆θ= ½(0+11)•3.251

∆θ= ½(11)•3.251

∆θ= 17.88rev.

5 0
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