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2 years ago

14

Calculate the number of free electrons and holes (in m-3) in an intrinsic semiconductor that has electron and hole mobilities of

0.35 and 0.17 m2/V-s, respectively, and a conductivity of 1.7 (Ω-m)-1. Use scientific notation.

1 answer:

allochka39001 [22]2 years ago

5 0

Answer:

ni = 2.04e19

Explanation:

we know that in semiconductor like intrinsic, when electron leave the band, it leave a hole in valence band so we have

n = p = ni

from intrinsic carrier concentration

$\sigma = n\left | e \right | \mu_e + n\left | e \right | \mu_h$

$\sigma = ni\left | e \right | \mu_e + ni\left | e \right | \mu_h$

$\sigma = ni \left | e \right | ( \mu_e + \mu_h)$

1.7 = ni * 1.6*10^{-19} * (.35 + .17)

ni = 2.014 *10^{19} m^{-3}

ni = 2.04e19

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The formula is x = 1/2 at^2 and I have managed to fill in the variables as this. d = 1/2 9.81 m/s^2 1^2

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Right, as you mentioned in the comments, you find $d$ by plugging in the different values of $t$ .

For $t=1\,\mathrm s$ , we have

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Similarly, for $t=2\,\mathrm s$ , you get

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3 years ago

you have been called to testify as a as an expert witness in a trial involving a head-on collision Car A weighs 1515 pounds and

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Answer:

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Explanation:

Car A mass= 1515 lb

Car B mass=1125 lb

Speed of car B is 46 miles/h

Distance before locking, d=19.5 ft

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Initial momentum of car B=mv where m is mass and v is velocity in ft/s

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$Momentum_B=1125*67.4666668 ft/s$

Initial momentum of car A is given by

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Taking East as positive and west as negative then the sum of initial momentum is

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The common velocity is represented as $v_c$ hence after collision, the final momentum is

$Momentum_final=(m_a+m_b)v_c=(1515+1125)v_c=2640v_c$

From the law of conservation of linear momentum, sum of initial and final momentum equals each other hence

$1515v_a-(1125*67.4666668 ft/s)= 2640v_c$

The acceleration of two cars $a=-\mug=-0.75*32.17=-24.1275 ft/s^{2}$

From kinematic equation

$v^{2}=u^{2}+2as$ hence

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

$0^{2}-(v_c)^{2}=2*-24.1275*19.5$

$v_c=\sqrt{2*24.1275*19.5}=30.67 ft/s$

Substituting the value of $v_c$ in equation $1515v_a-(1125*67.4666668 ft/s)= 2640v_c$

$1515v_a-(1125*67.4666668 ft/s)= 2640*30.67$

$1515v_a=(1125*67.4666668 ft/s)+2640*30.67$

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navik [9.2K]

An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. The minimum coefficient of static friction needed for this claim to be possible is 0.7

In an inclined plane, the coefficient of static friction is the angle at which an object slide over another.

As the angle rises, the gravitational force component surpasses the static friction force, as such, the object begins to slide.

Using the Newton second law;

$\sum F_x = \sum F_y = 0$

$\mathbf{mg sin \theta -f_s= N-mgcos \theta = 0 }$

So; On the L.H.S

$\mathbf{mg sin \theta =f_s}$

$\mathbf{mg sin \theta =\mu_s N}$

On the R.H.S

N = mg cos θ

Equating both force component together, we have:

$\mathbf{mg sin \theta =\mu_s \ mg \ cos \theta}$

$\mathbf{sin \theta =\mu_s \ \ cos \theta}$

$\mathbf{\mu_s = \dfrac{sin \theta }{ cos \theta}}$

From trigonometry rule:

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∴

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$\mathbf{\mu_s =\tan 35^0}}$

$\mathbf{\mu_s = 0.700}}$

Therefore, we can conclude that the minimum coefficient of static friction needed for this claim to be possible is 0.7

Learn more about static friction here:

brainly.com/question/24882156?referrer=searchResults

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