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Fantom [35]
3 years ago
5

You have a 2m long wire which you will make into a thin coil with N loops to generate a magnetic field of 3mT when the current i

n the wire is 1.2A. What is the radius of the coils and how many loops, N, are there
Physics
1 answer:
Anni [7]3 years ago
8 0

Answer:

<em>radius of the loop =  7.9 mm</em>

<em>number of turns N ≅ 399 turns</em>

Explanation:

length of wire L= 2 m

field strength B = 3 mT = 0.003 T

current I = 12 A

recall that field strength B = μnI

where n is the turn per unit length

vacuum permeability μ  = 4\pi *10^{-7}  T-m/A = 1.256 x 10^-6 T-m/A

imputing values, we have

0.003 = 1.256 x 10^−6 x n x 12

0.003 = 1.507 x 10^-5 x n

n = 199.07 turns per unit length

for a length of 2 m,

number of loop N = 2 x 199.07 = 398.14 ≅ <em>399 turns</em>

since  there are approximately 399 turns formed by the 2 m length of wire, it means that each loop is formed by 2/399 = 0.005 m of the wire.

this length is also equal to the circumference of each loop

the circumference of each loop = 2\pi r

0.005 = 2 x 3.142 x r

r = 0.005/6.284 = 7.9*10^{-4} m = 0.0079 m =<em> 7.9 mm</em>

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GuDViN [60]

Answer:

Ax = 0

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By = 8 sin phi = 8 sin (-34) = -4.47 m

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Ry = Ay + By = 6 - 4.47 = 1.53 m

R = (6.63^2 + 1.53^2)^1/2 = 6.80 m

tan theta = Ry / Rx = 1.53 / 6.8 = ,225

theta = 12.7 deg

7 0
3 years ago
I need help with some graph.<br><br> Which graph shows acceleration?
matrenka [14]

Answer:

The first graph

Explanation:

Graph A shows acceleration.

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2 years ago
Read 2 more answers
A crate with a mass of 110 kg glides through a space station with a speed of 4.0 m/s. An astronaut speeds it up by pushing on it
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Answer:

The final speed of the crate after the astronaut push to slow it down is 4.50 m/s

Explanation:

<u>Given:  </u>

The crate has mass m = 110 kg and an initial speed vi = 4 m/s.  

<u>Solution  </u>

We are asked to determine the final speed of the crate. We could apply the steps for energy principle update form as next  

Ef=Ei+W                                                 (1)

Where Ef and Ei are the find and initial energies of the crate (system) respectively. While W is the work done by the astronaut (surrounding).  

The system has two kinds of energy, the kinetic energy which associated with its motion and the rest energy where it has zero speed. The summation of both energies called the particle energy. So, equation (1) will be in the form  

(Kf + mc^2) = (KJ+ mc^2)                       (2)  

Where m is the mass of crate, c is the speed of light which equals 3 x 10^8 m/s and the term mc^2 represents the energy at rest and the term K is the kinetic energy.  

In this case, the rest energy doesn't change so we can cancel the rest energy in both sides and substitute with the approximate expression of the kinetic energy of the crate at low speeds where K = 1/2 mv^2 and equation (2) will be in the form

(1/2mvf^2+mc^2)=(1/2mvi^2 +mc^2)+W

1/2mvf^2=1/2mvi^2+W                              (3)

Now we want to calculate the work done on the crate to complete our calculations. Work is the amount of energy transfer between a source of an applied force and the object that experiences this force and equals the force times the displacement of the object. Therefore, the total work done will be given by  

W = FΔr                                                      (4)  

Where F is the force applied by the astronaut and equals 190 N and Δr is the displacement of the crate and equals 6 m. Now we can plug our values for F and Δr to get the work done by the astronaut  

W = F Δr= (190N)(6 m) = 1140 J  

Now we can plug our values for vi, m and W into equation (3) to get the final speed of the crate  

1/2mvf^2=1/2mvi^2+W

vf=5.82 m/s

This is the final speed of the first push when the astronaut applies a positive work done. Then, in the second push, he applies a negative work done on the crate to slow down its speed. Hence, in this case, we could consider the initial speed of the second process to be the final speed of the first process. So,  

vi' = vf

In this case, we will apply equation (3) for the second process to be in the

1/2mvf^2=1/2mvi'^2+W'                                 (3*)

The force in the second process is F = 170 N and the displacement is 4 m. The force and the displacement are in the opposite direction, hence the work done is negative and will be calculated by  

W'= —F Δr = —(170N)(4 m)= —680J

Now we can plug our values for vi' , m and W' into equation (3*) to get the final speed of the crate  

1/2mvf'^2=1/2mvi'^2+W'

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The final speed of the crate after the astronaut push to slow it down is 4.50 m/s

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3 years ago
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Answer:

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Magnification,  

M\ \alpha\ \dfrac{1}{f}

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Thus. Magnification will increase by decreasing the focal length.

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Shtirlitz [24]

Answer:

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Consequently, the reason for the acceleration month that the coefficient of kinetic friction is less than the coefficient of satic friction

3 0
3 years ago
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