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andriy [413]
3 years ago
14

. A solenoid coil consists of a single layer of 250 circular turns of wire with each turn having a 0.02m radius. The axial lengt

h of the coil is 0.3m. The coil is self-supporting, containing only air. (a) Determine the inductance of the coil, assuming that the magnetic field intensity is uniform inside the coil and zero elsewhere. (b) Find the stored energy in the magnetic field of the coil when the coil current is 18A.
Physics
1 answer:
Vinil7 [7]3 years ago
5 0

Answer:

a)    L = 3.29 10⁻⁴ H,  b)U = 5.33 10⁻²  J

Explanation:

a) The inductance is a solenoid this given carrier

           L = \frac{N \ \phi_B }{I}

The magnetic field inside the solenoid is

          B = μ₀ \frac{N}{l}  i

hence the magnetic flux

          Ф_B = B. A = μ₀ \frac{N \ A}{l \ i}

we substitute in the expression of inductance

          L = N² μ₀ A /l

let's find the area of ​​each turn

          A = π r²

         A = π 0.02²

         A = 1.2566 10⁻³ m²

let's calculate

          L = 250² 4π 10⁻⁷ 1.2566 10⁻² / 0.3

          L = 3.29 10⁻⁴ H

b) The stored energy is

           U = ½ L i²

let's calculate

            U = ½ 3.29 10⁻⁴ 18²

            U = 5.33 10⁻²  J

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malfutka [58]

Answer:

200 N/m

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0.31415 seconds

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Explanation:

m = Mass of glider = 0.5 kg

x = Displacement of spring

F = Force on spring = 6 N

From Hooke's law we have relation

F=kx\\\Rightarrow k=\frac{F}{x}\\\Rightarrow k=\frac{6}{0.03}\\\Rightarrow k=200\ N/m

The spring constant is 200 N/m

Angular frequency is given by

\omega=\sqrt{\frac{k}{m}}\\\Rightarrow \omega=\sqrt{\frac{200}{0.5}}\\\Rightarrow \omega=20\ rad/s

The angular frequency is 20 rad/s

Frequency is given by

f=\frac{\omega}{2\pi}\\\Rightarrow f=\frac{20}{2\pi}\\\Rightarrow f=3.18309\ Hz

The frequency is 3.18309 Hz

Time period is given by

T=\frac{1}{f}\\\Rightarrow T=\frac{1}{3.18309}\\\Rightarrow T=0.31415\ s

The time period is 0.31415 seconds

6 0
2 years ago
Why does the number 54,289.00 have seven significant figures and not five
nadya68 [22]
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3 years ago
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In World War II, there were several reported cases of airmen who jumped from their flaming airplanes with no parachute to escape
lina2011 [118]

Answer:

The deceleration of the pilot was  1.9 x 10⁴ m/s²

Explanation:

First, let´s calculate the velocity of the pilot 3 m above the ground. To do that, we need to know how much time the pilot was falling. The equation for the position and velocity of the pilot are as follows:

y = y0 + v0 * t + 1/2 * g * t²

v = v0 + g * t

where

y = position of the pilot at time t

y0 = initial position

v0 = initial velocity

t = time

v = velocity at time t

g = acceleration due to gravity

If we consider the ground as the center of our reference system and that the pilot fell with an initial velocity of 0 (the pilot would unlikely impulse himself to the ground), then the equations would be as follows:

y = y0 + 1/2 g * t²  

v = g * t

The time at which the pilot was 3.0 m above the ground will be:

3.0 m = 6000 m - 1/2 * 9.8 m/s² * t²

3.0 m - 6000 m / -4.9 m/s² = t²

t = 35.0 s

The velocity at that time will be:

v = -9.8 m/s² * 35.0 s = -343 m/s

After 35.0 s the pilot has a positive acceleration besides the acceleration due to gravity. Then, the equation for velocity and position will be:

v = v0 + a*t       now, v0 = -343 m/s and a ≠ g and a>0

y = y0 + v0 * t + 1/2 * a * t²    now, y0 = 3 m

Again, let´s find the time at which the pilot hits the ground:

v = v0 + a*t  

v-v0/ t  = a

Replacing in the equation for position:

y = y0 + v0 * t + 1/2 * ((v-v0)/t) * t²

y = y0 + v0 * t + 1/2 * v* t - 1/2 * v0 * t)

y = y0 + 1/2 v0 * t + 1/2 v * t

replacing with numbers:

0m = 3m + 1/2 * (-343 m/s)t + 1/2 *(-54 m/s)t

-3.0 m = - 198.5 m/s * t

t = -3m / -198.5 m/s

t =0.015 s

the upward acceleration was then:

v-v0/ t  = a

-54 m/s -(-343 m/s) / 0.015 s = a

a = 1.9 x 10⁴ m/s²

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When the north pole is tilted toward the sun, the southern hemisphere has the fewest hours of daylight due to winter season in it.

<h3>Which area has the fewest hours of daylight?</h3>

When the North Pole is tilted toward the sun, the Northern Hemisphere has the most hours of daylight while on the other hand, the Southern Hemisphere has the least hours of daylight.

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