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murzikaleks [220]

3 years ago

5

You are given a length (l) of wire that has radius (a)and are told to wind it into an inductor in the shape of a helix that has

a circular cross section of radius (r). The windings are to be as close together as possible without overlapping. Show that the self-inductance of this inductor is L = 1/4nrl/a.

1 answer:

LiRa [457]3 years ago

5 0

To solve this problem it is necessary to apply the concepts related to inductance on an inductor, which is mathematically described as:

$L = \mu_0 n^2 A l$

Where,

$\mu_0$ = Permeability constant (Described as 'n' at the problem equation)

l = Length

A = Cross-sectional Area

n = No of turn per unit length

The number of turns N is given by

$N = \frac{l}{2a}$

The number of turns per unit length n is

$n = \frac{N}{l} = \frac{1}{2a}$

The relationship of the cable lengths starts from assuming that the length 'a' is less than the length 'r', and therefore the length of the wire d would be related by:

$d = N(2\pi r)$

$d = \frac{l}{2a}2\pi r$

$d = \frac{\pi r}{a}l$

Solving to obtain l,

$l = \frac{ad}{\pi r}$

Substituting at the first equation,

$L = \mu_0 n^2 A l$

$L = \mu_0 (\frac{1}{2a})^2(\pi r^2)(\frac{ad}{\pi r})$

$L = \mu_0 (\frac{rd}{4a})$

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How many mm are in a micro-m?

Dahasolnce [82]

For this case we have that by definition, a micrometer is equivalent to a thousandth of a millimeter, that is, 0.001 millimeters.

Then, in other words, we have 0.001 millimeters in a micrometer.

Answer:

In a micrometer there are 0.001 millimeters.

4 0

3 years ago

A body is projected from the ground at an angle of 30° with the horizontal at an initial speed of 128 ft/s. Ignoring air frictio

Elanso [62]

Answer

given,

v = 128 ft/s

angle made with horizontal = 30°

now,

horizontal component of velocity

vx = v cos θ = 128 x cos 30° = 110.85 ft/s

vertical component of velocity

vy = v sin θ = 128 x sin 30° = 64 m/s

time taken to strike the ground

using equation of motion

v = u + at

0 =-64 -32 x t

t = 2 s

total time of flight is equal to

T = 2 t = 2 x 2 = 4 s

b) maximum height

using equation of motion

v² = u² + 2 a h

0 = 64² - 2 x 32 x h

64 h = 64²

h = 64 ft

c) range

R = v_x × time of flight

R = 110.85 × 4

R = 443.4 ft

4 0

3 years ago

You need to know the height of a tower, but darkness obscures the ceiling. You note that a pendulum extending from the ceiling a

Sonja [21]

Answer:

L=55.9m

Explanation:

The equation for the period of a simple pendulum is:

$T=2\pi\sqrt{\frac{L}{g}}$

In our case what we know is the period and the acceleration of gravity, and we need to know the length of the pendulum, so we can write:

$L=(\frac{T}{2\pi})^2g$

Which for our values is:

$L=(\frac{15s}{2\pi})^2(9.81m/s^2)=55.9m$

6 0

3 years ago

An object is originally moving at a constant velocity of 8 m/s in the -x direction. It moves at this constant velocity for 3 sec

aivan3 [116]

Answer:

244.64m

Explanation:

First, we find the distance traveled with constant velocity. It's simply multiplying velocity time the time that elapsed:

$x = V*t = -8\frac{m}{s} *3s = -24m$

After this, the ball will start traveling with a constant acceleration motion. Due to the fact that the acceleration is the opposite direction to the initial velocity, this motion will have 2 phases:

1. The velocity will start to decrease untill it reaches 0m/s.

2. Then, the velocity will start to increase at the rate of the acceleration.

The distance that the ball travels in the first phase can be found with the following expression:

$v^2 = v_0^2 + 2a*d$

Where v is the final velocity (0m/s), v_0 is the initial velocity (-8m/s) and a is the acceleration (+9m/s^2). We solve for d:

$d = \frac{v^2 - v_0^2}{2a} = \frac{(0m/s)^2 - (-8m/s)^2}{2*7m/s^2}= -4.57m$

Now, before finding the distance traveled in the second phase, we need to find the time that took for the velocity to reach 0:

$t_1 = \frac{v}{a} = \frac{8m/s}{7m/s^2} = 1.143 s$

Then, the time of the second phase will be:

$t_2 = 9s - t_1 = 9s - 1.143s = 7.857s$

Using this, we using the equations for constant acceleration motion in order to calculate the distance traveled in the second phase:

$x = \frac{1}{2}a*t^2 + v_0*t + x_0$

V_0, the initial velocity of the second phase, will be 0 as previously mentioned. X_0, the initial position, will be 0, for simplicity:

$x = \frac{1}{2}*7\frac{m}{s^2}*t^2 + 0m/s*t + 0m = 216.07m$

So, the total distance covered by this object in meters will be the sum of all the distances we found:

$x_total = 24m + 4.57m + 216.07m = 244.64m$

8 0

3 years ago

A mail carrier leaves the post ofﬁce and drives 2.00 km to the north. He then drives in a direction 60.0° south of east for 7.00

den301095 [7]

Answer:

$\theta=7^o$

Explanation:

<u>Displacement</u>

It is a vector that points to the final point where an object traveled from its starting point. If the object traveled to several points, then the individual displacements must be added as vectors.

The mail carrier leaves the post office and drives 2 km due north. The first displacement vector is

$\vec r_1=\ km$

Then the carrier drives 7 km in 60° south of east. The displacement has two components in the x and y axis given by

$\vec r_2=\ km=\ km$

Finally, he drives 9.5 km 35° north of east.

$\vec r_3=\ km=\ km$

The total displacement is

$\vec r_t=\ km+\ km+\ km$

$\vec r_t=\ km$

The direction can be calculated with

$\displaystyle tan\theta=\frac{1.39}{11.28}=0.1232$

$\boxed{\theta=7^o}$

7 0

3 years ago