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3 years ago

5

This person does work to stretch the rubber band. When will the potential energy be the greatest?

1 answer:

antiseptic1488 [7]3 years ago

7 0

the rubber band will be at its greatest when its fully stretched out

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In the steady state 1.2 ✕ 1018 electrons per second enter bulb 1. There are 6.3 ✕ 1028 mobile electrons per cubic meter in tungs

bekas [8.4K]

Answer:

E=12.2V/m

Explanation:

To solve this problem we must address the concepts of drift velocity. A drift velocity is the average velocity attained by charged particles, such as electrons, in a material due to an electric field.

The equation is given by,

$V=\frac{I}{nAq}$

Where,

V= Drift Velocity

I= Flow of current

n= number of electrons

q = charge of electron

A = cross-section area.

For this problem we know that there is a rate of 1.8*10^{18} electrons per second, that is

$\frac{I}{q} = 1.2*10^{18}$

$A= 1.3*10^{-8}m^2$

$n=6.3*10^{28} e/m^3$

$\omicron{O} = 1.2*10^{-4}(m/s)(N/c)$ Mobility

We can find the drift velocity replacing,

$V = \frac{1.2*10^{18}}{(1.3*10^{-8})(6.3*10^{28})}$

$V= 1.465*10^-3m/s$

The electric field is given by,

$E= \frac{V}{\omicron{O}}$

$E=\frac{1.465*10^-3}{1.2*10^{-4}}$

$E=12.2V/m$

7 0

3 years ago

A series circuit consists of a 100-ω resistor, a 10.0-μf capacitor, and a 0.350-h inductor. the circuit is connected to a 120-v

Tpy6a [65]

Current will be $I=\dfrac{V_{rms}}{Z}=\dfrac{V_{rms}}{\sqrt{ R^{2}+(X_{C}-X_{L})^{2}}}\\where~X_{c}=\dfrac{1}{j.\omega .C}~and~X_{L}=j.\omega.L~where~\omega=2.\pi f~and~f=60Hz$
now just pluf in the values and Voila..

7 0

2 years ago

A copper wire 20 m long and 4mm in diameter is attached to the Ceiling and a 400 N

Fiesta28 [93]

Answer:

A

Explanation:

5 0

1 year ago

What is the resistance of a light bulb if a potential difference of 120 V will produce a current of 0.5 A in the bulb?

Yakvenalex [24]

Explanation:

Ohm's law:

V = IR

120 V = (0.5 A) R

R = 240 Ω

6 0

3 years ago

Read 2 more answers

An electron is released from rest in a weak electric field given by E =-2.30 x 10-10 N/Cj. After the electron has traveled a ver

Goshia [24]

Explanation:

It is given that,

An electron is released from rest in a weak electric field of, $E=2.3\times 10^{-10}\ N/C$

Vertical distance covered, $s=1\ \mu m=10^{-6}\ m$

We need to find the speed of the electron. Let its speed is v. Using third equation of motion as :

$v^2-u^2=2as$

$v^2=2as$ .............(1)

Electric force is $F_e$ and force of gravity is $F_g$ . As both forces are acting in downward direction. So, total force is:

$F=mg+qE$

$F=9.1\times 10^{-31}\times 9.8+1.6\times 10^{-19}\times 2.3\times 10^{-10}$

$F=4.57\times 10^{-29}\ N$

Acceleration of the electron, $a=\dfrac{F}{m}$

$a=\dfrac{4.57\times 10^{-29}\ N}{9.1\times 10^{-31}\ kg}$

$a=50.21\ m/s^2$

Put the value of a in equation (1) as :

$v=\sqrt{2as}$

$v=\sqrt{2\times 50.21\times 10^{-6}}$

v = 0.010 m/s

So, the speed of the electron is 0.010 m/s. Hence, this is the required solution.

6 0

3 years ago