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Varvara68 [4.7K]
4 years ago
14

Interactive Solution 22.43 provides one model for solving this problem. The maximum strength of the earth's magnetic field is ab

out 6.9 x 10-5 T near the south magnetic pole. Suppose we want to utilize this field with a rotating coil to generate 54.1-Hz ac electricity. What is the minimum number of turns (area per turn = 0.024 m2) that the coil must have to produce an rms voltage of 170 V?
Physics
1 answer:
laila [671]4 years ago
5 0

Answer:

427097

Explanation:

V = Voltage = 170 V

B = Magnetic field = 6.9\times 10^{-5}\ T

f = Frequency = 54.1 Hz

A = Area = 0.024\ m^2

Number of turns is given by

N=\dfrac{V\sqrt{2}}{BA2\pi f}\\\Rightarrow N=\dfrac{170\sqrt{2}}{6.9\times 10^{-5}\times 0.024\times 2\pi\times 54.1}\\\Rightarrow N=427096.933536\approx 427097\ turns

Number of turns is 427097

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Answer:

(a) FN = 24.18 N

(b) a = 22.87 m/s²

Explanation:

Newton's second law of the  block:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Forces acting on the box

We define the x-axis in the direction parallel to the movement of the block on the surface   and the y-axis in the direction perpendicular to it.

F₁ : Horizontal force

F₂ : acting at an angle of θ to the horizontal,

W: Weight of the block  : In vertical direction

FN : Normal force : perpendicular to the direction the surface

fk : Friction force: parallel to the direction to the surface

Known data

m =3.1 kg : mass of the  block

F₁ = 65 N,  horizontal force

F₂ = 12.4 N acting at an angle of θ to the horizontal

θ = 30° angle θ of F₂ with respect to the horizontal

μk = 0.2 : coefficient of kinetic friction between the block and the surface

g = 9.8 m/s² : acceleration due to gravity

Calculated of the weight  of the block

W= m*g  = (3.1 kg)*(9.8 m/s²) = 30.38 N

x-y F₂ components

F₂x = F₂cos θ= (12.4)*cos(30)° = 10.74 N

F₂y = F₂sin θ= (12.4)*sin(30)° = 6.2 N

a)Calculated of the Normal force  (FN)

We apply the formula (1)

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FN+6.2-30.38 = 0

FN = -6.2+30.38

FN = 24.18 N

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fk=μk*N=  0.2* 24.18 N = 4.836 N

b) We apply the formula (1) to calculated acceleration of the block:

∑Fx = m*ax ,  ax= a  : acceleration of the block

F₁ + F₂x -fk = ( m)*a

65 N + 10.74 -4.836 = ( 3.1)*a

70.904 = ( 3.1)*a

a = (70.904 ) / ( 3.1)

a = 22.87 m/s²

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