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e-lub [12.9K]
3 years ago
11

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

ten. The cross-sectional area of the tungsten filament in bulb 1 is 1.3 ✕ 10-8 m2. The electron mobility in hot tungsten is 1.2 ✕ 10-4 (m/s)/(N/C). Calculate the magnitude of the electric field inside the tungsten filament in bulb 3.
Physics
1 answer:
bekas [8.4K]3 years ago
7 0

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

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What's an easy way to create an interference pattern of waves?
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3 years ago
Which of the following would illustrate a quadratic relation between the dependent and independent variables when graphed?
Kitty [74]

Answer: option A. a graph of the area of a circle vs. its radius r (A = πr²).



Explanation:



A quadratic relation between the dependent and independent variables shows the independent variable raised to the power of 2.



This is it is a polynomial with general form ax² + bx + c, whewre a, b, and c, named coeficients,  are constants.



The function is y =  ax² + bx + c, where x is the independent variable and y is the dependent variable.



As stated in the question, the area of a circle is given by A = πr².



In this case, A is the dependent variable and r is the independent variable.



π is assumed as the coefficient of the quadratic term, and the other coefficients are assumed 0, since there are no either terms on r or constants.



The equation a = 1/b  is an inverse relation, not a quadratic relation.



The relation of distance vs. time for a car moving at constant speed is a linear relation of the kind v = u + st.



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Therefore, the only quadratic relation is shown by  a graph of the area of a circle vs. its radius r.

3 0
2 years ago
A surgical microscope weighing 200 lb is hung from a ceiling by four springs with stiffness 25 lb/ft. The ceiling has a vibratio
Nikitich [7]

Answer:

If there is no damping, the amount of transmitted vibration that the microscope experienced is   = 5.676*10^{-3} \ mm

Explanation:

The motion of the ceiling is y = Y sinωt

y = 0.05 sin (2 π × 2) t

y = 0.05 sin 4 π t

K = 25 lb/ft  × 4  sorings

K = 100 lb/ft

Amplitude of the microscope  \frac{X}{Y}= [\frac{1+2 \epsilon (\omega/ W_n)^2}{(1-(\frac{\omega}{W_n})^2)^2+(2 \epsilon  \frac{\omega}{W_n})^2}]

where;

\epsilon = 0

W_n = \sqrt { \frac{k}{m}}

= \sqrt { \frac{100*32.2}{200}}

= 4.0124

replacing them into the above equation and making X the subject of the formula:

X = Y * \frac{1}{\sqrt{(1-(\frac{\omega}{W_n})^2)^2})}}

X = 0.05 * \frac{1}{\sqrt{(1-(\frac{4 \pi}{4.0124})^2)^2})}}

X = 5.676*10^{-3} \ mm

Therefore; If there is no damping, the amount of transmitted vibration that the microscope experienced is   = 5.676*10^{-3} \ mm

8 0
3 years ago
A top fuel dragster with a mass of 500.0 kg starts from rest and completes a quarter mile (402 m) race in a time of 5.0 s. The d
blsea [12.9K]

The average power needed to produce this final speed is 1069.1 hp.

Mass of the dragster,  m = 500.0 kg,

Displacement travelled by the dragster,  s = 402 m,

Time taken in this travel,  t = 5.0 s,

Final velocity of the dragster,  v = 130 m/s.

Let the initial velocity of the dragster be u and acceleration be a.

Using kinematical equation,  s = ut + (1/2)at^2.

402  =  u*5  + (1/2)*a*5^2

10*u + 25*a  = 804.      ...........(1)

Using kinematical equation, v = u +at.

130 = u + 5*a

5*u + 25*a = 650.       .............(2)

Solving (1)and (2), we get,

u =  30.8 m/s.

According to work-energy theorem,

Work done = change in kinetic energy

W  = (1/2)*m*(v^2 - u^2)

W = (1/2)*500*(130^2 - 30.8^2)

W  =  3987840. J

Therefore power rating of the dragster is given by,

P  ⇒  W/t. =  3987840/5 = 797568 watt.

P  ⇒ 797568/746 =  1069.1 hp.

Learn more about Power rating here brainly.com/question/20137708

#SPJ4

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