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barxatty [35]
3 years ago
12

A PMOS device with VT P = −1.2 V has a drain current iD = 0.5 mA when vSG = 3 V and vSD = 5 V. Calculate the drain current when:

(a) vSG = 2 V, vSD = 3 V; (b) vSG = 5 V, vSD = 2 V.
Engineering
1 answer:
Ksju [112]3 years ago
7 0
The answer is b ! Hope I helped
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A 1-mm-diameter methanol droplet takes 1 min for complete evaporation at atmospheric condition. What will be the time taken for
Svet_ta [14]

Answer:

Time taken by the 1\mu m diameter droplet is 60 ns

Solution:

As per the question:

Diameter of the droplet, d = 1 mm = 0.001 m

Radius of the droplet, R = 0.0005 m

Time taken for complete evaporation, t = 1 min = 60 s

Diameter of the smaller droplet, d' = 1\times 10^{- 6} m

Diameter of the smaller droplet, R' = 0.5\times 10^{- 6} m

Now,

Volume of the droplet, V = \frac{4}{3}\pi R^{3}

Volume of the smaller droplet, V' = \frac{4}{3}\pi R'^{3}

Volume of the droplet ∝ Time taken for complete evaporation

Thus

\frac{V}{V'} = \frac{t}{t'}

where

t' = taken taken by smaller droplet

\frac{\frac{4}{3}\pi R^{3}}{\frac{4}{3}\pi R'^{3}} = \frac{60}{t'}

\frac{\frac{4}{3}\pi 0.0005^{3}}{\frac{4}{3}\pi (0.5\times 10^{- 6})^{3}} = \frac{60}{t'}

t' = 60\times 10^{- 9} s = 60 ns

5 0
3 years ago
What is a p-n junction? Show by the diagram.
Natalija [7]

Answer:

The p-n junction is a region formed when a p -type semiconductor material is joined to an n-type semiconductor material

Explanation:

The p type semiconductor has holes as its majority charge carriers making it positively charged while the n –types has an overall negative charge. At the junction the holes move towards the electron until such a time when there is a balance in charges from both materials, which leads to the formation of the depletion zone as shown in the attachment below

                           

5 0
3 years ago
25 points and brainliest is it A, B, C, D
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Answer:

SIR IT IS D HOPE THIS HELPS (☞゚ヮ゚)☞☜(゚ヮ゚☜)

Explanation:

4 0
3 years ago
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A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 5.0 mm (0.20 in
Sonja [21]

The load is 17156 N.

<u>Explanation:</u>

First compute the flexural strength from:  

σ = FL / πR^{3}

   = 3000 \times (40 \times 10^-3) / π (5 \times 10^-3)^3

σ = 305 \times 10^6 N / m^2.

We can now determine the load using:

F = 2σd^3 / 3L

  = 2(305 \times 10^6) (15 \times 10^-3)^3 / 3(40 \times 10^-3)

F = 17156 N.  

7 0
3 years ago
The inner surface of a hollow cylinder is subjected to tangential and axial stresses of 40,000 and 24,000 psi, respectively. Det
Furkat [3]

Answer:

15,000 psi

Explanation:

The solution / solving is attach below.

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