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3 years ago

8

Ultra-thin semiconductor materials are of interest for future nanometer-scale transistors, but can present undesirably high resi

stance to current flow. How low must the resistivity of a semiconductor material be, to ensure that the resistance of a 2nm-thick, 10nm-long, 100nm-wide region does not exceed 100 ohms?

1 answer:

Mademuasel [1]3 years ago

6 0

Answer:

p = 2*10^(-7) ohm m

Explanation:

The resistivity and Resistance relationship is:

$p = \frac{R*A}{L}$

For lowest resistivity with R < 100 ohms.

We need to consider the possibility of current flowing across minimum Area and maximum Length.

So,

Amin = 2nm x 10 nm = 2 * 10^(-16) m^2

Lmax = 100nm

Using above relationship compute resistivity p:

$p = \frac{100*2*10^(-16)}{100*10^(-9)} \\\\p = 2 * 10^(-7)$

Answer: p = 2*10^(-7) ohm m

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A 20 dBm power source is connected to the input of a directional coupler having a coupling factor of 20 dB, a directivity of 35

lukranit [14]

Answer:

$P_O = 0.989 watt = 19.9 dBm$

Explanation:

Given data:

P_1 power = 20 dBm = 0.1 watt

coupling factor is 20dB

Directivity = 35 dB

We know that

coupling factor $= 10 log \frac{P_1}{P_f}$

solving for final power

$20 = 10 log\frac{P_1}{P_f}$

$2 = log \frac{P_1}{P_f}$

$100 = \frac{0.1}{P_f}$

$P_f = 0.001 watt = 0 dBm$

Directivity $D = 10 \frac{P_f}{P_b}$

$35 = 10 \frac{0.001}{P_b}$

$P_b = 3.162 \times 10^{-7} wattt$

output Power $= P_1 -P_f - P_b$

$= 0.1 - 0.001 - 3.162 \times 10^{-7}$

$P_O = 0.989 watt = 19.9 dBm$

6 0

4 years ago

Air is pumped from a vacuum chamber until the pressure drops to 3 torr. If the air temperature at the end of the pumping process

malfutka [58]

Answer:

The final pressure is 3.16 torr

Solution:

As per the question:

The reduced pressure after drop in it, P' = 3 torr = $3\times 0.133\ kPa$

At the end of pumping, temperature of air, $T = 5^{\circ}C = 278 K$

After the rise in the air temperature, $T' = 20^{\circ}C = 293 K$

Now, we know the ideal gas eqn:

PV = mRT

So

$P = \frac{m}{V}RT$

$P = \rho_{a}RT$ (1)

where

P = Pressure

V = Volume

$\rho_{a} = air\ density$

R = Rydberg's constant

T = Temperature

Using eqn (1):

$P = \rho_{a}RT$

$\rho_{a} = \frac{P}{RT}$

$\rho_{a} = \frac{3 times 0.133\times 10^{3}}{0.287\times 278} = 0.005 kg/m^{3}$

Now, at constant volume the final pressure, P' is given by:

$\frac{P}{T} = \frac{P'}{T'}$

$P' = \frac{P}{T}\times T'$

$P' = \frac{3}{278}\times 293 = 3.16 torr$

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vodka [1.7K]

Answer:

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5 0

3 years ago

Select the correct answer.

babunello [35]

Answer:

The information he needs next is;

B. Volume of paint needed per unit area

Explanation:

The operation Andy wants to perform = To apply paint on the walls of the house

The information Andy knows = The area of all walls in the house

The total volume of paint needed = The total area of the walls × The volume of paint needed for each unit area

Therefore, the information required is the volume of paint needed per unit area.

3 0

3 years ago

In a shear box test on sand a shearing force of 800 psf was applied with normal stress of 1750 psf. Find the major and minor pri

ryzh [129]

Answer:

The major and minor stresses are as 2060.59 psf, -310.59 psf and 1185.59 psf.

Explanation:

The major and minor principal stresses are given as follows:

$\sigma_{max}=\dfrac{\sigma_x+\sigma_y}{2}+\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}$

$\sigma_{min}=\dfrac{\sigma_x+\sigma_y}{2}-\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}$

Here

$\sigma_x$ is the normal stress which is 1750 psf
$\sigma_y$ is 0
$\tau_{xy}$ is the shear stress which is 800 psf

So the formula becomes

$\sigma_{max}=\dfrac{\sigma_x+\sigma_y}{2}+\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}\\\sigma_{max}=\dfrac{1750+0}{2}+\sqrt{\left(\dfrac{1750-0}{2}\right)^2+(800)^2}\\\sigma_{max}=875+\sqrt{\left(875)^2+(800)^2} \\\sigma_{max}=875+\sqrt{765625+640000}\\\sigma_{max}=875+1185.59\\\sigma_{max}=2060.59 \text{psf}$

Similarly, the minimum normal stress is given as

$\sigma_{min}=\dfrac{\sigma_x+\sigma_y}{2}-\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}\\\sigma_{min}=\dfrac{1700+0}{2}-\sqrt{\left(\dfrac{1700-0}{2}\right)^2+(800)^2}\\\sigma_{min}=875-\sqrt{(875)^2+(800)^2}\\\sigma_{min}=875-\sqrt{765625+640000}\\\sigma_{min}=875-1185.59\\\sigma_{min}=-310.59 \text{ psf}$

The maximum shear stress is given as

$\tau_{max}=\dfrac{\sigma_{max}-\sigma_{min}}{2}\\\tau_{max}=\dfrac{2060.59-(-310.59)}{2}\\\tau_{max}=\dfrac{2371.18}{2}\\\tau_{max}=1185.59 \text{psf}$

5 0

3 years ago