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hram777 [196]
3 years ago
12

A sinusoidally oscillating current I ( t ) with an amplitude of 9.55 A and a frequency of 359 cycles per second is carried by a

long, straight wire. A rectangular loop of copper wire with dimensions b = 72.2 cm by c = 32.5 cm is located a = 80.2 cm from the straight wire, and is coplanar with it. Calculate the average power P avg dissipated by the loop if its resistance is 64.3 Ω .
Physics
1 answer:
UNO [17]3 years ago
6 0

Answer:

P_{avg} = 6.283*10^{-9} \ W

Explanation:

Given that;

I₀ = 9.55 A

f = 359 cycles/s

b = 72.2 cm

c = 32.5 cm

a = 80.2 cm

Using the formula;

\phi = \frac{\mu_o Ic }{2 \pi} In (\frac{b+a}{b})

where;

E= \frac{d \phi}{dt}

E = \frac{\mu_o}{2 \pi}c In (\frac{b+a}{a}) I_o \omega cos \omega t

E_{rms} =   \frac { {\frac{\mu_o \ c}{2 \pi} In (\frac{b+a}{a}) I_o (2 \pi f)}}{\sqrt{2}}

Replacing our values into above equation; we have:

E_{rms} =   \frac { {\frac{4 \pi*10^{-7}*0.325}{2 \pi} In (\frac{72.2+80.2}{80.2}) *9.55 (2 \pi *359)}}{\sqrt{2}}

E_{rms} =   \frac {8.98909588*10^{-4} }{\sqrt{2}}

E_{rms} =   6.356*10^{-4} \ V

Then the P_{avg is calculated as:

P_{avg} = \frac{E^2}{R}

P_{avg} = \frac{(6.356*10^{-4})^2}{64.3}

P_{avg} = 6.283*10^{-9} \ W

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8 0
2 years ago
A ray of light passes from air into a block of clear plastic. How does the angle of incidence in the air compare to the angle of
andre [41]

Answer:

The angle of incidence is greater than the angle of refraction

Explanation:

Refraction occurs when a light wave passes through the boundary between two mediums.

When a ray of light is refracted, it changes speed and direction, according to Snell's Law:

n_1 sin \theta_1 = n_2 sin \theta_2

where :

n_1 is the index of refraction of the 1st medium

n_2 is the index of refraction of the 2nd medium

\theta_1 is the angle of incidence (the angle between the incident ray and the normal to the boundary)

\theta_2 is the angle of refraction (the angle between the refracted ray and the normal to the boundary)

In this problem, we have a ray of light passing from air into clear plastic. We have:

n_1=1.00 (index of refraction of air)

n_2=1.50 approx. (index of refraction in clear plastic)

Snell's Law can be rewritten as

sin \theta_2 =\frac{n_1}{n_2}sin \theta_1

And since n_2>n_1, we have

\frac{n_1}{n_2}

And so

\theta_2

Which means that

The angle of incidence is greater than the angle of refraction

6 0
3 years ago
In this example we see how thermal expansion can actually throw off the accuracy of a length-measuring device—namely, a tape mea
Ulleksa [173]

Answer:

The new length is 50.00885m

Explanation:

linear thermal expansion coefficient Fe 11.8e-6 /K

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∆L = Lα∆T = 50(11.8e-6)(35-20) = 0.00885 m

New length = ( 50.000 + 0.00885)m =

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5 0
3 years ago
Read 2 more answers
if you run around a circle at 4.5 m/s and the circle has a radius of 7.7 m, what is your centripetal acceleration?
madreJ [45]

Answer:

Centripetal acceleration,

a_{c} =2.63\ m/s^{2} }

Explanation:

Centripetal acceleration:

Centripetal acceleration is the idea that any object moving in a circle, in something called circular motion, will have an acceleration vector pointed towards the center of that circle.

Centripetal means towards the center.

Examples of centripetal acceleration (acceleration pointing towards the center of rotation) include such situations as cars moving on the cicular part of the road.

An acceleration is a change in velocity.

Formula for Centripetal acceleration:

a_{c} =\frac{(velocity)^{2} }{radius}

Given here,

Velocity = 4.5 m/s

radius = 7.7 m

To Find :

a_{c} = ?

Solution:

We have,

a_{c} =\frac{(velocity)^{2} }{radius}

Substituting  given value in it we get

a_{c} =\frac{(4.5)^{2}}{7.7} \\\\a_{c} =\frac{20.25}{7.7}\\\\a_{c} =2.629\ m/s^{2} \\\\\therefore a_{c} =2.63\ m/s^{2

Centripetal acceleration,

a_{c} =2.63\ m/s^{2} }

7 0
3 years ago
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