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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.

1 answer:

Ksju [112]3 years ago

7 0

The answer is b ! Hope I helped

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If a system of pulleys results in a force of 25% of the load, how far will the rope need to move to pull the load a distance of

GaryK [48]

Answer:

40 ft

Explanation:

Assuming no loss of energy in the system of pulleys, the work done is the same whether you move the load directly or through the pulleys.

W = Fd . . . . . . . . work is the product of force and distance

F(10 ft) = (0.25F)(d) . . . . . where d is the distance we want to find

d = 10F/(0.25F) = 40

The rope will need to move 40 feet.

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2 years ago

Whats the boolean expression of this circuit?

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Answer:

G8 = x0'x2' +x0'x3' +x1x2

Explanation:

The expression can be written different ways, depending on the need to avoid hazards. One of them is ...

$G_8=\overline{X_0}\,\overline{X_2}+\overline{X_0}\,\overline{X_3}+X_1X_2$

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A truth table and Karnaugh map are shown for the circuit. The terms used in the Boolean expression come from the corners, the upper half of the left- and right-columns, and the right half of the middle two rows. If a static hazard is to be avoided, a term x1x0' could be added representing the right column.

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2 years ago

A 15-mm-diameter steel bar is to be used as a torsion spring. If the torsional stress in the bar is not to exceed 110 MPa when o

just olya [345]

Answer:

The length of bar will be 2.82 m

Explanation:

Given that

d= 15 mm

r= 7.5 mm

Shear stress = 110 MPa

θ = 30° (30° = 30° x π/180° =0.523 rad)

θ = 0.523 rad

G for steel

G= 79.3 GPa

We know that

$\dfrac{\tau}{r}=\dfrac{G\theta }{L}$

$\dfrac{110}{7.5\times 10^{-3}}=\dfrac{79.3\times 10^3\times 0.523 }{L}$

L= 2. 82 m

The length of bar will be 2.82 m

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3 years ago

What happens to resistance in the strain gage and voltage drop from a connected Wheatstone bridge if you were to pull the strain

snow_tiger [21]

The resistance and voltage drop will still increase but at a smaller rate than the intended axis such as the long axis.

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3 years ago

The spherical pressure vessel has an inner diameter of 2 m and a thickness of 10 mm. A strain gauge having a length of 20 mm is

slava [35]

Question:

The spherical pressure vessel has an inner diameter of 2 m and a thickness of 10 mm. A strain gage having a length of 20 mm is attached to it, and it is observed to increase in length by 0.012 mm when the vessel is pressurized. Determine the pressure causing this deformation, and find the maximum in-plane shear stress, and the absolute maximum shear stress at a point on the outer surface of the vessel. The material is steel, for which Eₛₜ = 200 GPa and vₛₜ = 0.3.

Answer:

See explanation below

Explanation:

Given:

d = 2m = 2*10³ = 2000

thickness, t = 10 mm

Length of strain guage = 20 mm

i) Let's calculate d/t

$\frac{d}{t} = \frac{2000}{10} = 200$

Since $\frac{d}{t}$ is greater than length of strain guage, the pressure vessel is thin.

For the minimum normal stress, we have:

$\sigma max= \frac{pd}{4t}$

$\sigma max= \frac{2000p}{4 * 20}$

= 50p

For the minimum normal strain due to pressure, we have:

$E_max= \frac{change in L}{L_g}$

$= \frac{0.012}{20} = 0.60*10^-^3$

The minimum normal stress for a thin pressure vessel is 0.

$\sigma _min = 0$

i) Let's use Hookes law to calculate the pressure causing this deformation.

$E_max = \frac{1}{E} [\sigma _max - V(\sigma _initial + \sigma _min)]$

Substituting figures, we have:

$0.60*10^-^3 = \frac{1}{200*10^9} [50p - 0.3 (50p + 0)]$

$120 * 10^6 = 35p$

$p = \frac{120*10^6}{35}$

$p = 3.429 * 10^6$

p = 3.4 MPa

ii) Calculating the maximum in-plane shear stress, we have:

$\frac{\sigma _max - \sigma _int}{2}$

$= \frac{50p - 50p}{2} = 0$

Max in plane shear stress = 0

iii) To find the absolute maximum shear stress at a point on the outer surface of the vessel, we have:

$\frac{\sigma _max - \sigma _min}{2}$

$= \frac{50p - 0}{2} = 25p$

since p = 3.429 MPa

25p = 25 * 3.4 MPa

= 85.71 ≈ 85.7 MPa

The absolute maximum shear stress at a point on the outer surface of the vessel is 85.7 MPa

6 0

3 years ago