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2 years ago

9

An average hole drift velocity of 103 cm/sec results when 2 V is applied across a 1 cm long semiconductor bar. What is the hole

mobility inside the bar

1 answer:

Helga [31]2 years ago

8 0

Answer:

ε = 2 V/cm

Explanation:

To calculate the mobility inside this bar, we just need to apply the expression that let us determine the mobility. This expression is the following:

ε = ΔV / L

Where:

ε: Hole mobility inside the bar

ΔV: voltage applied in the bar

L: Length of the bar

We already have the voltage and the length so replacing in the above expression we have:

ε = 2 V / 1 cm

<h2>ε = 2 V/cm</h2><h2></h2>

The data of the speed can be used for further calculations, but in this part its not necessary.

Hope this helps

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An object is placed perpendicular to the the principal axis of convex lens of focal length 8,the distance of the object from the

Digiron [165]

Answer:

v = 24 cm and inverted image

Explanation:

Given that,

The focal length of the object, f = +8 cm

Object distance, u = -12 cm

We need to find the position &nature of the image. Let v be the image distance. Using lens formula to find it :

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

Put all the values,

$\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{8}+\dfrac{1}{(-12)}\\\\v=24\ cm$

So, the image distance from the lens is 24 cm.

Magnification,

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2 years ago

In an Atwood's machine, one block has a mass of 602.0 g, and the other a mass of 717.0 g. The pulley, which is mounted in horizo

Wittaler [7]

Answer:

The acceleration of the both masses is 0.0244 m/s².

Explanation:

Given that,

Mass of one block = 602.0 g

Mass of other block = 717.0 g

Radius = 1.70 cm

Height = 60.6 cm

Time = 7.00 s

Suppose we find the magnitude of the acceleration of the 602.0-g block

We need to calculate the acceleration

Using equation of motion

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t = time

a = acceleration

Put the value into the formula

$60.0\times10^{-2}=0+\dfrac{1}{2}\times a\times(7.00)^2$

$a=\dfrac{60.0\times10^{-2}\times2}{(7.00)^2}$

$a=0.0244\ m/s^2$

Hence, The acceleration of the both masses is 0.0244 m/s².

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3 years ago

Starting from rest, a 2.1x10^-4 kg flea springs straight upward. While the flea is pushing off from the ground, the ground exert

Rainbow [258]

Answer:

a) The flea's speed when it leaves the ground is $v=1.88m/s$

b) The flea move $18cm$ upward while it is pushing off

Explanation:

Hi

<u>Knwons</u>

Mass $m=2.1\times 10^{-4} kg$ , Work $W=3.7\times 10^{-4} J$ and Force $F=0.41N$

a) Here we are going to use $W=\frac{1}{2} mv^{2}$ , so $v=\sqrt{\frac{2W}{m} }= \sqrt{\frac{2(3.7\times 10^{-4} J)}{2.1\times 10^{-4} kg} }=1.88m/s$

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3 years ago

Two tuning forks of frequency 480 hz and 484 hz are struck simultaneously. what is the beat frequency resulting from the two sou

umka21 [38]

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Explanation:
The beat frequency is equal to the absolute value of the difference in frequency of the two waves. In other words, the number of beats per second is equal to the difference in frequency. It is due to the destructive and constructive interference. <span>According to this interference, sound will be soft or loud.

Hence. the formula is:
</span>Beat frequency = $f_b = |f_2 - f_1|$
<span>
Since,
</span> $f_1 = 480Hz$
$f_2 = 484Hz$

Therefore,
Beat frequency = $f_b = |484 - 480|$

=> Beat frequency = $f_b = 4Hz$
-i

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Answer:

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