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harina [27]
3 years ago
7

An electric generator contains a coil of 99 turns of wire, each forming a rectangular loop 73.9 cm by 34.9 cm. The coil is place

d entirely in a uniform magnetic field with magnitude B = 2.96 T and initially perpendicular to the coil's plane. What is in volts the maximum value of the emf produced when the loop is spun at 1200 rev/min about an axis perpendicular to the magnetic field?
Physics
2 answers:
serious [3.7K]3 years ago
8 0

Answer:

12078.46 V

Explanation:

Applying,

E₀ = BANω.................... Equation 1

Where: E₀ = maximum emf, B = magnetic Field, A = Area of the coil, N = Number of turns of the coil. ω = angular velocity

Given: N = 99 turns, B = 2.96 T, ω = 1200 rev/min = (1200+0.10472) = 125.664 rad/s

A = L×W, where L= Length = 93.9 cm = 0.939 m, W = width = 34.9 cm = 0.349 m

A = (0.939×0.349) = 0.328 m²

Substitute into equation 1

E₀ = 99(2.96)(0.328)(125.664)

E₀ = 12078.46 V

Hence the maximum value of the emf produced = 12078.46 V

Valentin [98]3 years ago
8 0

Answer:

The maximum emf induced in the loop is 9498.268 V

Explanation:

Given;

number of turns of coil, N = 99 turns

area of the rectangular loop, A = 0.739 m x 0.349 m = 0.2579 m²

magnetic field strength, B = 2.96 T

angular speed of the loop, ω = 1200 rev/min

angular speed of the loop, ω (rad/s) = (2π x 1200) / 60 = 125.68 rad/s

The maximum value of the emf produced is calculated using the formula below;

ξ = NABω

Substitute the given values and calculate the maximum emf induced;

ξ = (99)(0.2579)(2.96)(125.68)

ξ = 9498.268 volts

Therefore, the maximum emf induced in the loop is 9498.268 V

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A box of mass 50 kg is pushed hard enough to set it in motion across a flat surface. Then a 99-N pushing force is needed to keep
vitfil [10]

Answer:

0.20

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F-\mu mg=0\\\mu = \frac{F}{mg}=\frac{99 N}{(50 kg)(9.8 m/s^2)}=0.20

7 0
3 years ago
A 34 kg bowling ball with a radius of 22 cm starts from rest at the top of an incline 4.4 m in height. Find the translational sp
liberstina [14]

Answer:

7.85 m/s.

Explanation:

Given,

Mass of the bowling ball, m = 34 Kg

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height of the inclination, h = 4.4 m

transnational velocity = ?

Moment of inertia of bowling ball,

I = \dfrac{2}{5}mr^2

Using conservation of energy

mgh = \dfrac{1}{2}I\omega^2 + \dfrac{1}{2}mv^2

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v =\sqrt{\dfrac{gh}{0.7}}

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Speed of the bowling ball is equal to 7.85 m/s.

3 0
3 years ago
Distance travelled by a free falling object in the first second is: a) 4.9m b) 9.8m c) 19.6m d) 10m​
zvonat [6]
  • Time=1s=t
  • Acceleration due to gravity=g=9.8m/s^2
  • Distance=s

In free fall

\boxed{\sf s=-\dfrac{1}{2}gt^2}

\\ \sf\longmapsto s=-\dfrac{1}{2}\times 9.8(1)^2

\\ \sf\longmapsto s=-4.9(1)

\\ \sf\longmapsto s=-4.9m

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\\ \sf\longmapsto s=4.9m

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6 0
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