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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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GuDViN [60]

Answer:

ΔE = 8.77 × 10¹¹ J

Explanation:

given,

²¹⁴₈₄Po -----> ²¹⁰₈₂Pb + 42 He

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He-4 = 4.00260 amu

1 kg = 6.022 × 10²⁶ amu;

NA = 6.022 × 10²³ mol⁻¹

c = 2.99792458 × 10⁸ m/s

energy of molecule using equation

ΔE = Δm c²

Δm is mass difference and c is speed of light

Δm = 209.98284 + 4.00260 - 213.99519

Δm = - 0.00975 amu

1 amu = 1.66 x 10⁻²⁷ kg

- 0.00975 amu = - 0.00975 x 1.66 x 10⁻²⁷ Kg

                         = -0.016185 x 10⁻²⁷ Kg

total mass = 6.022 × 10²³ x -0.016185 x 10⁻²⁷

                 = -0.097467 x 10⁻⁴ Kg

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8 0
2 years ago
At a local swimming pool, the diving board is elevated h = 5.5 m above the pool's surface and overhangs the pool edge by L = 2 m
Margaret [11]

Answer:

Part a)

t = \sqrt{\frac{2h}{g}}

Part b)

t = 1.06 s

Part c)

L  = 4.86 m

Explanation:

Part a)

The height of the diving board is given as

h = 5.5 m

now the speed of the diver is given as

v_0 = 2.7 m/s

when the diver will jump into the water then his displacement in vertical direction is same as that of height of diving board

So we will have

y = v_y t + \frac{1}{2}at^2

h = 0 + \frac{1}{2}gt^2

t = \sqrt{\frac{2h}{g}}

Part b)

t = \sqrt{\frac{2h}{g}}

plug in the values in the above equation

t = \sqrt{\frac{2(5.5 m)}{9.81}

t = 1.06 s

Part c)

Horizontal distance moved by the diver is given as

d = v_0 t

d = 2.7 \times 1.06

d = 2.86 m

so the distance from the edge of the pool is given as

L = 2.86 + 2

L  = 4.86 m

4 0
2 years ago
Military specifications often call for electronic devices to be able to withstand accelerations of 10 g. to make sure that their
trapecia [35]
The solution for this problem is:
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= sqrt(2d/a), = sqrt (98.1 m/sec^2/0.094m) = 32.3050619 sec per cycle

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= 5.14 Hz would be the answer
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2 years ago
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gogolik [260]
False is the correct answer
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2 years ago
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AleksAgata [21]

Answer:

Explanation:

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40.48 L = \sqrt{\frac{3.92}{7.25\times10^{-7}} }

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2 years ago
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