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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 large well-mixed tank of unknown volume, open to the atmosphere initially, contains pure water. The initial height of the solu
trasher [3.6K]

Answer:

The exact time when the sample was taken is = 0.4167337 hr

Explanation:

The diagram of a sketch of the tank is shown on the first uploaded image

Let A denote the  first inlet

Let B denote the second inlet

Let C denote the single outflow from the tank

From the question we are given that the diameter of A is = 1 cm = 0.01 m

                              Area of  A is  = \frac{\pi}{4}(0.01)^{2} m^{2}

                                                    = 7.85 *10^{-5}m^{2}

Velocity of liquid through A = 0.2 m/s

  The rate at which the liquid would flow through the first inlet in terms of volume  = \frac{Volume of Inlet }{time} = Velocity * Area i.e is m^{2} * \frac{m}{s}   = \frac{m^{3}}{s}

             = 0.2 *7.85*10^{-5} \frac{m^{3}}{s}

  The rate at which the liquid would flow through the first inlet in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume

                              =  1039.8 * 0.2 * 7.85 *10^{-5} Kg/s

                              = 0.016324 \frac{Kg}{s}

From the question the diameter of B = 2 cm = 0.02 m

                                           Area of B = \frac{\pi}{4} * (0.02)^{2} m^{2} = 3.14 * 10^{-4}m^{2}

                                     Velocity of liquid through B = 0.01 m/s

The rate at which the liquid would flow through the first inlet in terms of volume  = \frac{Volume of Inlet }{time} = Velocity * Area i.e is m^{2} * \frac{m}{s}   = \frac{m^{3}}{s}

             = 3.14*10^{-4} *0.01 \frac{m^{3}}{s}

The rate at which the liquid would flow through the second inlet in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume

                              = 1053 * 3.14*10^{-6} \frac{Kg}{s}

                              = 0.00330642 \frac{Kg}{s}

From the question The flow rate in term of volume of the outflow at the time of measurement is given as  = 0.5 L/s

And also from the question the mass of  potassium chloride  at the time of measurement is given as 13 g/L

So The rate at which the liquid would flow through the outflow in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume

                              = 13\frac{g}{L} * 0.5 \frac{L}{s}

                              =  \frac{6.5}{1000}\frac{Kg}{s}       Note (1 Kg = 1000 g)

                              = 0.0065 kg/s

Considering potassium chloride

         Let denote the  rate at which liquid flows in terms of mass as   as \frac{dm}{dt} i.e change in mass with respect to time hence

           Input(in terms of mass flow ) - output(in terms of mass flow ) = Accumulation in the Tank(in terms of mass flow )

         

      (0.016324 + 0.00330642) - 0.0065 = \frac{dm}{dt}

          \int\limits {\frac{dm}{dt} } \, dx  =\int\limits {0.01313122} \, dx

      => 0.01313122 t = (m - m_{o})

  From the question  (m - m_{o})  is given as = 19.7 Kg

Hence the time when the sample was taken is given as

               0.01313122 t = 19.7 Kg

      =>  t = 1500.2414 sec

            t = .4167337 hours (1 hour = 3600 seconds)

5 0
3 years ago
Can somebody help me with that
skelet666 [1.2K]

Answer:

I think it's 23 ohms.

Explanation:

Not entirely sure about it.

hope this helps

4 0
2 years ago
If link AB of the four-bar linkage has a constant counterclockwise angular velocity of 58 rad/s during an interval which include
katrin2010 [14]

Answer:Vb=-6i-(-0.1ωab+8)j m/s

Explanation:

Va=V0+Va0

Va=V0+(ra0 x ωao)

ω=Angular velocity of link A0

Using r0a=0.1m;

Va=V0+(0.1i x ω0a K)

Va=0

ixk=j

Va=0+0.1ω0aj

Calculating te velocity of using te equation below

Vb=Va+Vba

Vb=Va+ωab x rba

ωab=40rad/s

rab=-0.21i+0.15j

Va=0.1ω0aj

Vb=Va+ωabxrba

Vb=0.1ω0aj+40k x -(0.21i+0.15j)

Vb=0.1ω0aj-8j-6i

Vb=-6i-(-0.1ωab+8)j m/s

5 0
3 years ago
Which of the following is not a function of the suspension system?_____
tatyana61 [14]
The awnser is D cause it is
4 0
3 years ago
An air conditioner removes heat steadily from a house at a rate of 750 kJ/min while drawing electric power at a rate of 6 kW. De
Paraphin [41]

Answer:

a. 2.08, b. 1110 kJ/min

Explanation:

The power consumption and the cooling rate of an air conditioner are given. The COP or Coefficient of Performance and the rate of heat rejection are to be determined. <u>Assume that the air conditioner operates steadily.</u>

a. The coefficient of performance of the air conditioner (refrigerator) is determined from its definition, which is

COP(r) = Q(L)/W(net in), where Q(L) is the rate of heat removed and W(net in) is the work done to remove said heat

COP(r) = (750 kJ/min/6 kW) x (1 kW/60kJ/min) = 2.08

The COP of this air conditioner is 2.08.

b. The rate of heat discharged to the outside air is determined from the energy balance.

Q(H) = Q(L) + W(net in)

Q(H) = 750 kJ/min + 6 x 60 kJ/min = 1110 kJ/min

The rate of heat transfer to the outside air is 1110 kJ for every minute.

5 0
3 years ago
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