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3 years ago

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A quantum well is a region of semiconductor or metal film that confines particles to a small region, resulting in quantized ener

gies that are useful for the operation of a variety of electronic devices. If an electron in a quantum well is confined to an area that is 510 nm wide, what is the uncertainty in the electron's velocity along the same direction?

1 answer:

murzikaleks [220]3 years ago

5 0

Answer:

226.2 m/sec

Explanation:

We have given $\Delta x=510nm=510\times 106{-9}m$

The plank's constant $h=6.67\times 10^{-34}$

Mass of electron $m=9.11\times 10^{-31}kg$

Now according to Heisenberg uncertainty principle $\Delta v\geq \frac{h}{2\pi m\Delta x}$

So $\Delta v\geq \frac{6.6\times 10^{-34}}{2\times 3.14\times 9.11\times 10^{-31}\times 510\times 10^{-9}}=226.2m/sec$

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Technician A says the counter gear is actually several gears machined out of a single piece of steel. Technician B says the coun

dexar [7]

Answer:

Option c. (Both Technician A and B are correct)

Explanation:

A transmission system consists of 3 shafts. The input shaft, the counter shaft, and the main shaft. The clutch gear always rotates with input shaft and is a crucial element of the input shaft.

The counter shaft is actually several gears machined out of a single piece of steel. The counter shaft may also be called counter gear or cluster gear. It is a secondary shaft that runs parallel to the mainshaft in a gearbox and is used to provide powers to machine components such as the drive axle.

The main gears (also called the speed gears) on main shaft (also known as the output shaft) are used to transfer rotation from counter shaft to the output shaft.

Hence in the light of above description, both technician A and B are correct.

4 0

4 years ago

Read 2 more answers

Two long, straight wires are parallel and 20 cm apart. One carries a current of 2.2 A, the other a current of 5.9 A. (a) If the

Leya [2.2K]

Answer: $1.298\times 10^{-5}\ N/m$

Explanation:

Given

Current in the first wire $I_1=2.2\ A$

Current in the second wire $I_2=5.9\ A$

wires are $20\ cm$ apart

Force per unit length between the current-carrying wires is

$\Rightarrow \dfrac{F}{l}=\dfrac{\mu_oI_1I_2}{2\pi r}$

Force exerted by the wires is the same

Put the values

$\Rightarrow \frac{F}{l}=f=\dfrac{4\pi \times 10^{-7}\times 2.2\times 5.9}{2\pi \times 0.2}=1.298\times 10^{-5}\ N/m$

This force will be repulsive in nature as the current is flowing opposite

4 0

3 years ago

A 0.106-A current is charging a capacitor that has square plates 4.60 cm on each side. The plate separation is 4.00 mm.

nikdorinn [45]

Answer:

a

$\frac{d \phi_{E}}{dt} =1.1977 *10^{10} \ V\cdot m/s$

b

$I = 0.106 \ A$

Explanation:

From the question we are told that

The current is $I = 0.106 \ A$

The length of one side of the square $a = 4.60 \ cm = 0.046 \ m$

The separation between the plate is $d = 4.0 mm = 0.004 \ m$

Generally electric flux is mathematically represented as

$\phi_E = \frac{Q}{\epsilon_o}$

differentiating both sides with respect to t is

$\frac{d \phi_{E}}{dt} = \frac{1}{\epsilon_o} * \frac{d Q}{ dt}$

=> $\frac{d \phi_{E}}{dt} = \frac{1}{\epsilon_o} *I$

Here $\epsilon_o$ is the permitivity of free space with value

$\epsilon _o = 8.85*10^{-12} C/(V \cdot m)$

=> $\frac{d \phi_{E}}{dt} = \frac{0.106}{8.85*10^{-12}}$

=> $\frac{d \phi_{E}}{dt} =1.1977 *10^{10} \ V\cdot m/s$

Generally the displacement current between the plates in A

$I = 8.85*10^{-12} * 1.1977 *10^{10}$

=> $I = 0.106 \ A$

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3 years ago

Two balls of mass m1 and m2, with velocities v1 and v2 collide head on. Is there any way for both balls to have zero velocity af

nordsb [41]

Answer:

Explanation:

As the final Kinetic energy is zero or less than initial kinetic energy, the collision must be inelastic.

In Inelastic collision both the bodies must stick together as final velocity is zero for both the bodies.

To conserve the momentum, momentum associated before the collision of first must be equal and opposite to the momentum associated with the second ball.

i.e.

$m_1v_1=m_2v_2$

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3 years ago

A speed-time graph is shown below:

alexandr1967 [171]

Answer:

<em>The average acceleration of the object is 0.5 cm/s^2</em>. Answer: A)

Explanation:

<u>Average Acceleration</u>

The average acceleration of a moving object is the rate at which the speed changed over time in a specific period. It can be calculated as follows:

$\displaystyle \bar a=\frac{v_f-v_o}{t}$

Where vf is the final speed, vo is the initial speed and t is the time.

The speed-time shows the speed starts at 0 cm/s and ends at 4 cm/s over the 8 seconds period. The average acceleration is:

$\displaystyle \bar a=\frac{4-0}{8}=0.5$

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A) Correct. The average acceleration is 0.5 cm/s^2

B) Incorrect. The average acceleration is 0.5 cm/s^2

C) Incorrect. The average acceleration is 0.5 cm/s^2

D) Incorrect. The average acceleration is 0.5 cm/s^2

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3 years ago