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3 years ago

Which is a dopant for a p-type semiconductor? arsenic indium phosphorus antimony

Physics

1 answer:

erastova [34]3 years ago

4 0

Answer:

As opposed to n-type semiconductors, p-type semiconductors have a larger hole concentration than electron concentration.

Explanation:

In p-type semiconductors, holes are the majority carriers and electrons are the minority carriers. A common p-type dopant for silicon is boron or gallium. hope this you :)

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During the spin cycle the time dependent angular speed of a washing machine drum is given by the equation ω(t) = at + bt2 - ct4

anastassius [24]

Answer:

α = 2.6 +1.7 t - 0.14 t³

Explanation:

Given that

ω(t) = at + bt² - ct⁴

where a = 2.6 rad/s², b = 0.85 rad/s³ and c = 0.035 rad/s⁵

We know that angular acceleration is the rate of change of angular velocity

α = dω/dt

ω(t) = at + bt² - ct⁴

dω/dt= a + 2 b t - 4 ct³

α = a + 2 b t - 4 ct³

By putting the values of a b and c

α = a + 2 b t - 4 ct³

α = 2.6 + 2 x 0.85 t - 4 x 0.035 t³

α = 2.6 +1.7 t - 0.14 t³

5 0

3 years ago

Read 2 more answers

How long does it take a car travelling at 80 m/s to decelerate to 18 m/s with an acceleration rate of -16m/s?

sertanlavr [38]

initial velocity=u=80m/s
Final velocity=v=18m/s
Acceleration=a=-16m/s^2
Time=t

We know

$\boxed{\sf a=\dfrac{v-u}{t}}$

$\\ \sf\longmapsto t=\dfrac{v-u}{a}$

$\\ \sf\longmapsto t=\dfrac{18-80}{-16}$

$\\ \sf\longmapsto t=\dfrac{-62}{16}$

$\\ \sf\longmapsto t=3.8s$

4 0

3 years ago

A small block has constant acceleration as it slides down a frictionless incline. The block is released from rest at the top of

Alchen [17]

Answer:

The speed of the block when it is 5.00 m from the top of the incline is 3.04 m/s

Explanation:

given information:

s = 7.80 m

v = 3.8 m/s

if s = 5 m, v?

first we have to find the acceleration of the block using the following equation:

v² = v₀² + 2as, v₀² = 0 thus

3.8² = 2 a 7.8

a = 0.93

so, if s = 5m the final speed is

v² = 2 (0.93) (5)

= 9.26

v = √9.26

= 3.04 m/s

5 0

4 years ago

The Doppler effect using ultrasonic waves of frequency 2.25 × 106 Hz is used to monitor the heartbeat of a fetus. A (maximum) be

miss Akunina [59]

Answer:

$v = 2.88 \times 10^7 m/s$

Explanation:

As per Doppler's effect we know that the frequency of the sound that is observed by the detector is the reflected sound

This reflected sound is given as

$f' = f (\frac{v - v_h}{v + v_h})$

$f' = 2.25\times 10^6(\frac{v - 1540}{v + 1540})$

so we know that the beat frequency is

$\Delta f = 240 Hz$

so we will have

$f - f' = \Delta f$

$2.25 \times 10^6 - 2.25\times 10^6(\frac{v - 1540}{v + 1540}) = 240$

$1 - (\frac{v - 1540}{v + 1540}) = 1.07 \times 10^{-4}$

so we have

$0.99989 = (\frac{v - 1540}{v + 1540})$

$1.99989\times 1540 = 1.067 \times 10^{-4} v$

$v = 2.88 \times 10^7 m/s$

4 0

4 years ago

What is the prmary source of energy inside of the earth

FromTheMoon [43]

The correct answer is the energy of the sun

8 0

2 years ago

Read 2 more answers