What device is similar to a sandwich that has layers of semiconductors

A current of 0.4 A flows through a wire. How many electrons flow through a cross section of

Free_Kalibri [48]

9 × 10²¹ electrons flow through a cross section of the wire in one hour.

<h3>What is the relation between current and charge?</h3>

Mathematically, current = charge / time
In S.I. unit, Charge is written in Coulomb and time in second.

<h3>What is the amount of charge flown through a wire for one hour if it carries 0.4 A current?</h3>

Charge= current × time
Current= 0.4 A, time = 1 hour= 3600 s
Charge= 0.4× 3600

= 1440 C

<h3>How many numbers of electrons present in 1440C of charge?</h3>

One electron= 1.6 × 10^(-19) C
So, 1440 C = 1440/1.6 × 10^(-19)

= 9 × 10²¹ electrons

Thus, we can conclude that the 9 × 10²¹ electrons flow through a cross section of the wire in one hour.

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4 0

2 years ago

Consider a horizontal layer of the dam wall of thickness dx located a distance x above the reservoir floor. What is the magnitud

adoni [48]

Answer:

Explanation:

Attached is the solution

6 0

3 years ago

When a certain string is clamped at both ends, the lowest four resonant frequencies are 50, 100, 150, and 200 Hz. When the strin

mario62 [17]

Answer:

Explanation:

Given

Lowest four resonance frequencies are given with magnitude

50,100,150 and 200 Hz

The frequency of vibrating string is given by

$f=\frac{n}{2L}\sqrt{\frac{T}{\mu }}$

where n=1,2,3 or ...n

L=Length of string

T=Tension

$\mu =$ Mass per unit length

When string is clamped at mid-point

Effecting length becomes $L'=0.5 L$

Thus new Frequency becomes

$f' =\frac{n}{L}\sqrt{\frac{T}{\mu }}$

i.e. New frequency is double of old

so new lowest four resonant frequencies are 100,200,300 and 400 Hz

4 0

2 years ago

If a player through a basketball to the target with an initial velocity of 17 m/s making an angle of 30 degrees with the horizon

Svetllana [295]

Answer:

The final position made with the vertical is 2.77 m.

Explanation:

Given;

initial velocity of the ball, V = 17 m/s

angle of projection, θ = 30⁰

time of motion, t = 1.3 s

The vertical component of the velocity is calculated as;

$V_y = Vsin \theta\\\\V_y = 17 \times sin(30)\\\\V_y = 8.5 \ m/s$

The final position made with the vertical (Yf) after 1.3 seconds is calculated as;

$Y_f = V_yt - \frac{1}{2}g t^2\\\\Y_f = (8.5 \times 1.3 ) - (\frac{1}{2} \times 9.8 \times 1.3^2)\\\\Y_f = 11.05 \ - \ 8.281\\\\Y_f = 2.77 \ m$

Therefore, the final position made with the vertical is 2.77 m.

3 0

2 years ago

Which situation best describes the act of reducing?

yawa3891 [41]

A - i think
paying bills online?

5 0

3 years ago

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