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xenn [34]
2 years ago
15

A technician is working on an MRI machine. To test it, the technician turns on the MRI machine that produces a strong magnetic f

ield, but then he accidentally drops a screwdriver in this magnetic field. As the screwdriver falls, an emf is induced across the length of the metal shaft, L, of the screwdriver. At one instant, the screwdriver falls with a speed of 2.1 m/s and is oriented as shown in figure so that the induced emf is maximized. The magnitude of the magnetic field at this instant is 2.5 T, and the length of the metal shaft, L, is 10 cm. What is the induced emf across the length of the metal shank at this instant?
Physics
1 answer:
AnnZ [28]2 years ago
4 0

Answer:

0.525 V

Explanation:

Noting the given data in the question:

we have,

The speed of the fall of the screw driver, v = 2.1 m/s

The magnitude of the magnetic field, B = 2.5 T

The length of the metal shaft of the screw driver, L = 10 cm = 0.1 m

Now, the induce emf (E) is given as:

E = B × L × v

on substituting the values in the above equation, we get

E = 2.5 × 0.1 × 2.1

or

E = 0.525 V

Hence, the emf induced across the length of the metal shank will be 0.525 V

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To start the analysis of this circuit you must write energy conservation (loop) equations. Each equation must involve a round-tr
REY [17]

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Explanation:

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2 years ago
A solid cylinder of mass 12.0 kg and radius 0.250 m is free to rotate without friction around its central axis. If you do 75.0 J
faltersainse [42]

Answer:

20 rad/s

Explanation:

mass, m = 12 kg

radius, r = 0.250 m

Moment of inertia of cylinder, I = 1/2 mr²

I = 0.5 x 12 x 0.250 x 0.250 = 0.375 kgm^2

Work done = Change in kinetic energy

Initial K = 0

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W = 1/2 Iω²

ω² = 2W/ I = 2 x 75 / (0.375)

ω = 20 rad/s

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8 0
2 years ago
A car with speed v and an identical car with speed 2v both travel the same circular section of an unbanked road. If the friction
yawa3891 [41]

Answer:

F'=\dfrac{F}{4}

Explanation:

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For faster car on the road,

F=\dfrac{mv^2}{r}

v = 2v

F=\dfrac{m(2v)^2}{r}

F=4\dfrac{m(v)^2}{r}..........(1)

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F'=\dfrac{mv^2}{r}............(2)

Equation (1) becomes,

F=4F'

F'=\dfrac{F}{4}

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2 years ago
Two window washers, Bob and Joe, are on a 3.00 m long, 395 N scaffold supported by two cables attached to its ends. Bob weighs 8
WINSTONCH [101]

Answer:

- the forces on the left hand side is 1.038 kN

- the forces on the right hand side is 1.483 kN

Explanation:

Given the data in the question, as illustrated in the image below;

Length of the scaffold = 3 m

weight of the scaffold = 395 N

Weight of Bob = 805 N and stands 1 m from the left end

weight of washing equipment = 500N and on sits 2 m from the left end

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so the force on the left cable will be;

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T_{left =  \frac{1}{3m}[ 1610 + 592.5 + 500 + 410 ]

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Therefore, the forces on the right hand side is 1.483 kN

5 0
1 year ago
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