Ask question

Ask question

All categories

3 years ago

15

A device is turned on and 3.20 A flows through it 0.130 ms later. What is the self-inductance of the device (in mH) if an induce

d 140 V emf opposes this

1 answer:

jeka57 [31]3 years ago

8 0

Answer:

the self-inductance of the device is 5.69 mH

Explanation:

Given;

change in current, ΔI = 3.2 A

change in time, Δt = 0.13 ms = 0.13 x 10⁻³ s

induced emf, E = 140

The self-inductance is calculated as;

$E = L\frac{\Delta I}{\Delta t} \\\\where;\\\\L \ is \ the \ self-inductance\\\\\L = \frac{E\Delta t}{\Delta I} \\\\L =\frac{140 \times 0.13 \times 10^{-3}}{3.2} \\\\L = 0.00569 \ H\\\\L = 5.69 \ mH$

Therefore, the self-inductance of the device is 5.69 mH

You might be interested in

Astronaut X of mass 50kg floats next to Astronaut Y of mass 100kg while in space, as shown in the figure. The positive direction

jonny [76]

Answer:

C

Explanation:

The change in momentum of x has to be the opposite of the change in momentum of Y because the momentum is just transferred from one to another. But I'm still trying to figure it out how to calculate.

5 0

2 years ago

The electron gun in a television tube accelerates electrons (mass = 9.11x10^-31 kg, charge = 1.6 x 10^-19C) from rest to 3 x10^7

stich3 [128]

Answer:

Electric field acting on the electron is 127500 N/C.

Explanation:

It is given that,

Mass of an electron, $m=9.11\times 10^{-31}\ kg$

Charge on electron, $q=1.6\times 10^{-19}\ C$

Initial speed of electron, u = 0

Final speed of electron, $v=3\times 10^7\ m/s$

Distance covered, s = 2 cm = 0.02 m

We need to find the electric field required. Firstly, we will find the acceleration of the electron from third equation of motion as :

$a=\dfrac{v^2-u^2}{2s}$

$a=\dfrac{(3\times 10^7)^2-0}{2\times 0.02}$

$a=2.25\times 10^{16}\ m/s^2$

According to Newton's law, force acting on the electron is given by :

F = ma

$F=9.1\times 10^{-31}\times 2.25\times 10^{16}$

$F=2.04\times 10^{-14}\ N$

Electric force is given by :

F = q E, E = electric field

$E=\dfrac{F}{q}$

$E=\dfrac{2.04\times 10^{-14}}{1.6\times 10^{-19}}$

E = 127500 N/C

So, the electric field is 127500 N/C. Hence, this is the required solution.

8 0

3 years ago

What happens to kinetic energy when you increase the mass?

Rus_ich [418]

Increasing mass increases kinetic energy. This can be seen in the equation KE = 1/2 (m) (v)^2

If you found this helpful, please brainliest me!

5 0

3 years ago

The Greeks noticed that some stars were larger and moved haphazardly. What were these stars?

Varvara68 [4.7K]

Answer:

i guess it's shooting star

5 0

3 years ago

One string of a certain musical instrument is 70.0 cm long and has a mass of 8.79 g . It is being played in a room where the spe

Svetach [21]

To solve this problem we will apply the concepts of linear mass density, and the expression of the wavelength with which we can find the frequency of the string. With these values it will be possible to find the voltage value. Later we will apply concepts related to harmonic waves in order to find the fundamental frequency.

The linear mass density is given as,

$\mu = \frac{m}{l}$

$\mu = \frac{8.79*10^{-3}}{70*10^{-2}}$

$\mu = 0.01255kg/m$

The expression for the wavelength of the standing wave for the second overtone is

$\lambda = \frac{2}{3} l$

Replacing we have

$\lambda = \frac{2}{3} (70*10^{-2})$

$\lambda = 0.466m$

The frequency of the sound wave is

$f_s = \frac{v}{\lambda_s}$

$f_s = \frac{344}{0.768}$

$f_s = 448Hz$

Now the velocity of the wave would be

$v = f_s \lambda$

$v = (448)(0.466)$

$v = 208.768m/s$

The expression that relates the velocity of the wave, tension on the string and linear mass density is

$v = \sqrt{\frac{T}{\mu}}$

$v^2 = \frac{T}{\mu}$

$T= \mu v^2$

$T = (0.01255kg/m)(208.768m/s)^2$

$T = 547N$

The tension in the string is 547N

PART B) The relation between the fundamental frequency and the $n^{th}$ harmonic frequency is

$f_n = nf_1$

Overtone is the resonant frequency above the fundamental frequency. The second overtone is the second resonant frequency after the fundamental frequency. Therefore

$n=3$

Then,

$f_3 = 3f_1$

Rearranging to find the fundamental frequency

$f_1 = \frac{f_3}{3}$

$f_1 = \frac{448Hz}{3}$

$f_1 = 149.9Hz$

7 0

3 years ago