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3 years ago

All MOS devices are subject to damage from:________

Engineering

1 answer:

Alchen [17]3 years ago

6 0

Answer:

d. all of these

Explanation:

Electrostatic discharge will generally produce excess voltage in a local area that results in excessive current and excessive heat. It will blast a crater in an MOS device, or melt bond wires, or cause damage of other sorts. In short, MOS devices are subject to damage from "all of these."

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A cylindrical specimen of some metal alloy having an elastic modulus of 124 GPa and an original cross-sectional diameter of 4.2

IrinaVladis [17]

Answer:

the maximum length of the specimen before deformation is 0.4366 m

Explanation:

Given the data in the question;

Elastic modulus E = 124 GPa = 124 × 10⁹ Nm⁻²

cross-sectional diameter D = 4.2 mm = 4.2 × 10⁻³ m

tensile load F = 1810 N

maximum allowable elongation Δl = 0.46 mm = 0.46 × 10⁻³ m

Now to calculate the maximum length $l$ for the deformation, we use the following relation;

$l$ = [ Δl × E × π × D² ] / 4F

so we substitute our values into the formula

$l$ = [ (0.46 × 10⁻³) × (124 × 10⁹) × π × (4.2 × 10⁻³)² ] / ( 4 × 1810 )

$l$ = 3161.025289 / 7240

$l$ = 0.4366 m

Therefore, the maximum length of the specimen before deformation is 0.4366 m

5 0

3 years ago

Note that common skills are listed toward the top, and less common skills are listed toward the bottom.

Sladkaya [172]

Answer:

BDEG

Explanation:

got it right on the test on edge because i used my b r a i n

5 0

3 years ago

A furnace wall composed of 200 mm, of fire brick. 120 mm common brick 50mm 80% magnesia and 3mm of steel plate on the outside. I

Liula [17]

Answer:

fire brick / common brick : 1218 °C
common brick / magnesia : 1019 °C
magnesia / steel : 90.06 °C
heat loss: 4644 kJ/m^2/h

Explanation:

The thermal resistance (R) of a layer of thickness d given in °C·m²·h/kJ is ...

R = d/k

so the thermal resistances of the layers of furnace wall are ...

R₁ = 0.200/4 = 0.05 °C·m²·h/kJ

R₂ = 0.120 2.8 = 3/70 °C·m²·h/kJ

R₃ = 0.05/0.25 = 0.2 °C·m²·h/kJ

R₄ = 0.003/240 = 1.25×10⁻⁵ °C·m²·h/kJ

So, the total thermal resistance is ...

R₁ +R₂ +R₃ +R₄ = R ≈ 0.29286 °C·m²·h/kJ

The rate of heat loss is ΔT/R = (1450 -90)/0.29286 = 4643.70 kJ/(m²·h)

The temperature drops across the various layers will be found by multiplying this heat rate by the thermal resistance for the layer:

fire brick: (4543.79 kJ/(m²·h))(0.05 °C·m²·h/kJ) = 232 °C

so, the fire brick interface temperature at the common brick is ...

1450 -232 = 1218 °C

For the next layers, the interface temperatures are ...

common brick to magnesia = 1218 °C - (3/70)(4643.7) = 1019 °C

magnesia to steel = 1019 °C -0.2(4643.7) = 90.06 °C

_____

<em>Comment on temperatures</em>

Most temperatures are rounded to the nearest degree. We wanted to show the small temperature drop across the steel plate, so we showed the inside boundary temperature to enough digits to give the idea of the magnitude of that.

5 0

3 years ago

When your fixing a car, what is the first thing you want to do?

Shtirlitz [24]

Answer:

Changing oil.

Explanation:

You need to regularly check and change your car’s oil to ensure smooth running of the vehicle and to prolong the lifespan of its engine.

6 0

2 years ago

Marks

Lina20 [59]

Answer:

i think C

Explanation:

if not really sorry

3 0

2 years ago