The thermal energy is carried by electromagnetic waves

A particular NMOS device has parameters VT N = 0.6 V, L = 0.8µm, tox = 200 Å, and µn = 600 cm2 /V–s. A drain current of ID = 1.2

NeTakaya

Answer:

$W= 3.22 \mu m$

Explanation:

the transistor In saturation drain current region is given by:

$i_D}=K_a(V_{GS}-V_{IN})^2$

Making $K_a$ the subject of the formula; we have:

$K_a=\frac {i_D} {(V_{GS} - V_{IN})^2}$

where;

$i_D = 1.2m$

$V_{GS}= 3.0V$

$V_{TN} = 0.6 V$

$K_a=\frac {1.2m} {(3.0 - 0.6)^2}$

$K_a = 208.3 \mu A/V^2$

Also;

$k'_n}=\frac{\mu n (\frac{cm^2}{V-s} ) \epsilon _{ox}(\frac{F}{cm} ) }{t_{ox}(cm)}$

where:

$\mu n (\frac{cm^2}{V-s} ) = 600$

$\epsilon _{ox}=3.9*8.85*10^{-14}$

${t_{ox}(cm)=200*10^{-8}$

substituting our values; we have:

$k'_n}=\frac{(600)(3.988.85*10^{-14})}{(200*10^{-8})}$

$k'_n}=103.545 \mu A/V^2$

Finally, the width can be calculated by using the formula:

$W= \frac{2LK_n}{k'n}$

where;

L = $0.8 \mu m$

$W= \frac{2*0.8 \mu m *208.3 \mu}{103.545 \mu}$

$W= 3.22 \mu m$

4 0

4 years ago

A 3.7 g mass is released from rest at C which has a height of 1.1 m above the base of a loop-the-loop and a radius of 0.2 m . Th

Dmitrij [34]

Answer:

Normal force = 0.326N

Explanation:

Given that:

mass released from rest at C = 3.7 g = 3.7 × 10⁻³ kg

height of the mass = 1.1 m

radius = 0.2 m

acceleration due to gravity = 9.8 m/s²

We are to determine the normal force pressing on the track at A.

To to that;

Let consider the conservation of energy relation; which says:

mgh = mgr + 1/2 mv²

gh = gr + 1/2 v²

gh - gr = 1/2v²

g(h-r) = 1/2v²

v² = 2g(h-r)

However; the normal force will result to a centripetal force; as such, using the relation

N =mv²/r

replacing the value for v² = 2g(h-r) in the above relation; we have:

Normal force = 2mg(h-r)/r

Normal force = 2 × 3.7 × 10⁻³ × 9.8 ( 1.1 - 0.2 )/ 0.2

Normal force = 0.065268/0.2

Normal force = 0.32634 N

Normal force = 0.326N

6 0

3 years ago

Ammonia enters the expansion valve of a refrigeration system at a pressure of 10 bar and a temperature of 20oC and exits at 3.0

icang [17]

Answer:

Ammonia enters the expansion valve of a refrigeration system at a pressure of 10 bar and a temperature of 20 o C and exits at 5.0 bar. The refrigerant undergoes a throttling process. Determine the temperature, in o C, and the quality of the refrigerant at the exit of the expansion valve.

Explanation:

3 0

3 years ago

What are the characteristics of Polyamide?

Tems11 [23]

The two most important polyamides are poly(hexamethylene adipamide) (Nylon 6,6) and polycaprolactam (Nylon 6). Both have excellent mechanical properties including high tensile strength, high flexibility, good resilience, low creep and high impact strength (toughness).

7 0

3 years ago

when removing the pistons and rods assemblies from a cylinder block technician a positions the throw of the crankshaft at the to

defon

Answer: Technician A

Reason: Piston and rod are connected with crank shaft and connecting rod. Smaller end of connecting rod is connected with piston and bigger end is connected with crankshaft.

<u>Explanation: </u>

When removing the piston and rod assemblies from a cylinder block: The technician with correct approach. So to remove piston from cylinder technician must throw crankshaft first and the remove connecting rod by losing nuts and caps...So technician A is in right way. Technician B using hammer to remove piston from the rod not possible because connecting rod and piston connected by nuts and caps can’t be separate by hammer.

Technician A positions the throw of the crankshaft at the top of its stroke and removes the connecting rod nuts and cap.

Technician B covers the rod bolts with hammers and pushes the piston and rod assembly out with the wooden hammer handle or wooden drift and supports the piston as it comes out of the cylinder.

4 0

3 years ago