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sergij07 [2.7K]
2 years ago
15

What does the current in a semi-conductor is produced by?

Engineering
2 answers:
Ulleksa [173]2 years ago
6 0
Current flow in a semiconductor arises from the motion of charge carriers in both the conduction and valence bands. As explained in chapter 4, the mobile charges in the conduction band are electrons and those in the valence band are holes.
lesya [120]2 years ago
5 0
Current flow in a semiconductor arises from the motion of charge carriers in both the conduction and valence bands. As explained in chapter 4, the mobile charges in the conduction band are electrons and those in the valence band are holes.

Here u goo!!
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AC motor characteristics require the applied voltage to be proportionally adjusted by an AC drive whenever the frequency is chan
Margarita [4]
The answer is false
6 0
2 years ago
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What does the word “robot” mean? A.Clone B. Athlete C. Servant D. Actor
hram777 [196]

Answer:

a. clone

Explanation:

4 0
2 years ago
Calculate the rate at which body heat is conducted through the clothing of a skier in a steady- state process, given the followi
olga2289 [7]

Answer:

230.4W

Explanation:

Heat transfer by conduction consists of the transport of energy in the form of heat through solids, in this case a jacket.

the equation is as follows

Q=\frac{KA(T2-T1)}{L} \\

Where

Q=heat

k=conductivity=0.04

A=Area=1.8m^2

T2=33C

T1=1C

L=thickness=1cm=0.01mQ=\frac{(0.04)(1.8m^2)(33-1)}{0.01m}

Q=230.4W

the skier loses heat at the rate of 230.4W

4 0
3 years ago
At the instant under consideration, the hydraulic cylinder AB has a length L = 0.75 m, and this length is momentarily increasing
Inessa [10]

Answer:

vB = - 0.176 m/s   (↓-)

Explanation:

Given

(AB) = 0.75 m

(AB)' = 0.2 m/s

vA = 0.6 m/s

θ = 35°

vB = ?

We use the formulas

Sin θ = Sin 35° = (OA)/(AB) ⇒  (OA) = Sin 35°*(AB)

⇒   (OA) = Sin 35°*(0.75 m) = 0.43 m

Cos θ = Cos 35° = (OB)/(AB) ⇒  (OB) = Cos 35°*(AB)

⇒   (OB) = Cos 35°*(0.75 m) = 0.614 m

We apply Pythagoras' theorem as follows

(AB)² = (OA)² + (OB)²

We derive the equation

2*(AB)*(AB)' = 2*(OA)*vA + 2*(OB)*vB

⇒  (AB)*(AB)' = (OA)*vA + (OB)*vB

⇒  vB = ((AB)*(AB)' - (OA)*vA) / (OB)

then we have

⇒  vB = ((0.75 m)*(0.2 m/s) - (0.43 m)*(0.6 m/s) / (0.614 m)

⇒  vB = - 0.176 m/s   (↓-)

The pic can show the question.

7 0
3 years ago
Read 2 more answers
How many electrons move past a fixed reference point every t = 2.55 ps if the current is i = 7.3 μA ? Express your answer as an
iris [78.8K]

Answer:

116.3 electrons

Explanation:

Data provided in the question:

Time, t = 2.55 ps = 2.55 × 10⁻¹² s

Current, i = 7.3 μA = 7.3 × 10⁻⁶ A

Now,

we know,

Charge, Q = it

thus,

Q = (7.3 × 10⁻⁶) × (2.55 × 10⁻¹²)

or

Q = 18.615 × 10⁻¹⁸ C

Also,

We know

Charge of 1 electron, q = 1.6 × 10⁻¹⁹ C

Therefore,

Number of electrons past a fixed point = Q ÷ q

= [ 18.615 × 10⁻¹⁸ ] ÷ [ 1.6 × 10⁻¹⁹ ]

= 116.3 electrons

4 0
2 years ago
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