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3 years ago

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. A cathode ray tube (CRT) is a device that produces a focused beam of electrons in a vacuum. The electrons strike a phosphor-co

ated glass screen at the end of the tube, which produces a bright spot of light. The position of the bright spot of light on the screen can be adjusted by deflecting the electrons with electrical fields, magnetic fields, or both. Although the CRT tube was once commonly found in televisions, computer displays, and oscilloscopes, newer appliances use a liquid crystal display (LCD) or plasma screen. You still may come across a CRT in your study of science. Consider a CRT with an electron beam average current of 25.00μA25.00μA . How many electrons strike the screen every minute?

1 answer:

charle [14.2K]3 years ago

8 0

Answer:

The value is $n= 9.375 *10^{15} \ electrons$

Explanation:

From the question we are told that

The average current is $I = 25.0 \mu A = 25.0 *10^{-6} \ A$

Generally the quantity of charge (electron ) is mathematically represented as

$Q = ne$

Here e is the charge on a single electron with value $e = 1.60 *10^{-19} \ C$

Generally current is mathematically represented as

$I = \frac{Q}{t}$

=> $I = \frac{ne}{t}$

Here t is time which is given as 1 minutes = 60 seconds

and n is the number of electrons

So

$25.0 *10^{-6} = \frac{ n* 1.60 *10^{-19}}{60}$

=> $60 * 25.0 *10^{-6} = n* 1.60 *10^{-19}$

=> $n= \frac{60 * 25.0 *10^{-6} }{ 1.60 *10^{-19} }$

=> $n= 9.375 *10^{15} \ electrons$

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Quentin is playing baseball and is batting. He swings and hits the ball sending it flying to the outfield. Which is greater- the

Bess [88]

Answer:

The force of the ball on the bat is same as the force of the bat on the ball.

Explanation:

A bat hits the ball and the ball moves to the out filed.

According to the Newton's third law, for every action there is an equal and opposite reaction.

The action and the reaction forces acts on the two different bodies but the magnitude of the force is same.

As the ball is hitted by the bat, the bat exerts the force on the ball and the same force is exerted on the bat by the ball according to the Newton's third law.

So, the force of the ball on the bat is same as the force of the bat on the ball but the direction of force is opposite.

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3 years ago

Problem #3: 3.6. A diode has wdo = 0.4 µm and φj = 0.85 V. (a) What reverse bias is required to triple the depletion-layer width

Alchen [17]

Answer:

a) 6.8 Volt

b) 1.21Цm

Explanation:

We are given from the question that

The zero -bias depletion layer width( $W_{do}$ ) is 0.4Цm

The built in voltage φj is 0.85V

Now to calculate the reverse voltage( $V_{R}$ ) that would be required to triple the depletion - layer width.

The depletion - layer width ( $W_{d}$ ) of the diode has the formula

$W_{d} = W_{do} \sqrt{1 + \frac{V_{R} }{Qj} }$

For three times of $W_{d}$ we have

$3W_{d} = W_{do} \sqrt{1 +\frac{V_{R} }{Qj} }$

=> $\frac{V_{R} }{Qj} = 3^{2} -1$

=> $V_{R} = 8Qj$

Substituting value of φj

We have

$V_{R} = 8(0.85V)$

= 6.8 V

The required bias voltage $V_{R}$ is 6.8 V

The solution for the b part of the question is uploaded on first image

7 0

3 years ago

At what distance from a long, straight wire carrying a current of 8.7 A is the magnetic field due to the wire equal to the stren

deff fn [24]

Answer:

r= 3.2 cm

Explanation:

Given that

I= 8.7 A

B= 5.4 x 10⁻⁵ T

μo=1.25664 x 10⁻⁶

We know that magnetic filed in wire at a distance r given as

$B=\dfrac{\mu_oI}{2\pi r}$

$r=\dfrac{\mu_oI}{2\pi B}$

By putting the values

$r=\dfrac{1.25664\times 10^{-6}\times 8.7}{2\times \pi \times 5.4\times 10^{-5}}\ m$

r=0.032 m

r= 3.2 cm

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3 years ago

A series circuit that is connected to a 50 V, 60 Hz source is made up of 25 ohm resistor, capacite wieh X= 18 ohms, and inductor

Allisa [31]

Answer:

the impedance of the circuit is 25.7 ohms.

Explanation:

It is given that,

Voltage, V = 50 volts

Frequency, f = 60 Hz

Resistance, R = 25 ohms

Capacitive resistance, $X_C=18\ ohms$

Inductive resistance, $X_L=24\ ohms$

We need to find the impedance of the circuit. It is given by :

$Z=\sqrt{R^2+(X_L-X_C)^2}$

$Z=\sqrt{25^2+(24-18)^2}$

Z = 25.7 ohms

So, the impedance of the circuit is 25.7 ohms. Hence, this is the required solution.

6 0

3 years ago

A straight fin is made from copper (k = 388 W/m-K) and is 0.5 cm in diameter and 30 cm long. The temperature at the base of the

erastovalidia [21]

Answer:

The rate of transfer of heat is 0.119 W

Solution:

As per the question:

Diameter of the fin, D = 0.5 cm = 0.005 m

Length of the fin, l =30 cm = 0.3 m

Base temperature, $T_{b} = 75^{\circ}C$

Air temperature, $T_{infty} = 20^{\circ}$

k = 388 W/mK

h = $20\ W/m^{2}K$

Now,

Perimeter of the fin, p = $\pi D = 0.005\pi \ m$

Cross-sectional area of the fin, A = $\frac{\pi}{4}D^{2}$

A = $\frac{\pi}{4}(0.5\times 10^{-2})^{2} = 6.25\times 10^{- 6}\pi \ m^{2}$

To calculate the heat transfer rate:

$Q_{f} = \sqrt{hkpA}tanh(ml)(T_{b} - T_{infty})$

where

$m = \sqrt{\frac{hp}{kA}} = \sqrt{\frac{20\times 0.005\pi}{388\times 6.25\times 10^{- 6}\pi}} = 41.237$

Now,

$Q_{f} = \sqrt{20\times 388\times 0.005\pi\times 6.25\times 10^{- 6}\pi}tanh(41.237\times 0.3)(75 - 20) = 0.119\ W$

5 0

3 years ago