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Over [174]
3 years ago
6

A CIRCULAR COIL OF 20 TURNS OF RADIUS 10.0 cm IS PERPENDICULAR TO A 0.30 tesla MAGNETIC FIELD. HOW MUCH INDUCED emf IS CREATED W

HEN THE COIL IS ROTATED IN 0.12 s, SO ITS FACE IS PARALLEL TO THE FIELD?
Physics
1 answer:
attashe74 [19]3 years ago
6 0

Answer:

1.57 V

Explanation:

The induced emf ε = -ΔΦ/Δt where Φ = magnetic flux = NABcosθ where N = number of turns of circular coil = 20, A = area of coil = πr² where r = radius of coil = 10.0 cm = 0.1 m , B = strength  of magnetic field = 0.30 T,θ = angle between normal to are and magnetic field and Δt = change in time = 0.12 s

Now ΔΦ = change in magnetic flux through circular coil = Φ' - Φ

Φ = magnetic field when A is perpendicular to B , that is θ = 90°. So, Φ = NABcos90 = 0 and Φ' = magnetic field when A is parallel to B , that is θ = 0°. So, Φ' = NABcos0 = NAB

So, ΔФ = Ф' - Ф

= 0 - NAB

= -NAB

= -Nπr²B

= -20 × π(0.1 m)² × 0.30 T

= -0.19 Wb

So, ε = -ΔΦ/Δt

= -(-0.19 Wb)/0.12 s

= 1.57 Wb/s

= 1.57 V

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3 years ago
A 15.0-μF capacitor is charged by a 130.0-V power supply, then disconnected from the power and connected in series with a 0.280-
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The resonant frequency of a circuit is the frequency \omega_0 at which the equivalent impedance of a circuit is purely real (the imaginary part is null).

Mathematically this frequency is described as

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Where

L = Inductance

C = Capacitance

Our values are given as

C = 15*10^{-6}\mu F

L = 0.280*10^{-3}mH

Replacing we have,

f = \frac{1}{2\pi}(\sqrt{\frac{1}{LC}})

f = \frac{1}{2\pi}(\sqrt{\frac{1}{(15*10^{-6})(0.280*10^{-3})}})

f= 2455.81Hz

From this relationship we can also appreciate that the resonance frequency infers the maximum related transfer in the system and that therefore given an input a maximum output is obtained.

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7 0
4 years ago
A bicyclist of mass 60kg supplies 340W of power while riding into a 5 m/s head wind. The frontal area of the cyclist and bicycle
NikAS [45]

Answer:

87.1 mph

Explanation:

We are given that

Mass,m=60 kg

Power,P=340 W

Speed,v=5 m/s

Area,A=0.344 m^2

Drag coefficient,C_d=0.88

Coefficient of rolling resistance,\mu_r=0.007

Friction force,f=\mu_rmg=0.007\times 60\times 9.8=4.1 N

Where g=9.8 m/s^2

Let speed of cyclist=v'

Drag force,F_d=\frac{1}{2}\rho_{air}AC_dv^2

Density of air,\rho_{air}=1.225 kg/m^3

F_d=\frac{1}{2}\times 1.225\times 0.344\times 0.88(5)^2=4.635N

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v'=\frac{340}{8.735}=38.9m/s

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1 m=0.00062137 miles

1 hour=3600 s

7 0
4 years ago
Shearing of the wool is done with special instruments called_____​
Kobotan [32]

Answer:

The machine used is called a squaring shear, power shear, or guillotine.

Explanation:

5 0
3 years ago
Read 2 more answers
Could anyone help with this? :)
rjkz [21]
I think its d- theory
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3 years ago
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