How will the delay and active power per device change as you increase the doping density of both the N- and the P-MOSFET?

Answer: the discharge observed at a v-notch weir is 66.7 in³/s

Explanation:

Given that;

Notch angle ∅ = 90°

height above the weir is 3 inches { head + head correction factor) h + k = 3 in

Discharge Q = ?

To determine the discharge observed, we us the following expression

Q = 4.28Ctan(∅/2) ( h + k )^5/2

where Q is discharge, C is discharge coefficient, ∅ is notch angle, h is head and k is head correction factor

now we substitute

Q = 4.28 × 1 × tan(90/2) ( 3 )^5/2

Q = 4.28 × 1 × 1 × 15.5884

Q =  66.7 in³/s

Therefore the discharge observed at a v-notch weir is 66.7 in³/s

Answer:

For 31 P NMR spectra

low temperature

there is two types of 19f seen in low temperature and they are

therefore in low temperature the 31p couples with the two types of 19F seen ( [tex]b_{f} and c_{f}[/tex]to form a triplet and this couples more with [tex]a_{f}[/tex] to form a doublet. i.e. one (1) peak

High temperature

At High temperature The exchange is fast here therefore the 31p spectra sees all 19p at once and in the same environment leading to the formation of one (1) peak

For 19 P NMR spectra

low temperature

In low temperature [tex]a_{f}, b_{f} , c_{f}[/tex] is fixed  and the environment where [tex]b_{f} and c _{f}[/tex] is the same hence a peak is formed and another peak is formed by [tex]a_{f}[/tex] that makes the number of peaks = 2 peaks

High temperature

In high temperature [tex]a_{f}, b_{f} , c_{f}[/tex]  exchange very fast therefore one peak is formed for all, since the fast exchanges makes NMR machine to take an average and produce just one peak for all

Explanation:

For 31 P NMR spectra

low temperature

there is two types of 19f seen in low temperature and they are

therefore in low temperature the 31p couples with the two types of 19F seen ( [tex]b_{f} and c_{f}[/tex]to form a triplet and this couples more with [tex]a_{f}[/tex] to form a doublet. i.e. one (1) peak

High temperature

At High temperature The exchange is fast here therefore the 31p spectra sees all 19p at once and in the same environment leading to the formation of one (1) peak

For 19 P NMR spectra

low temperature

In low temperature [tex]a_{f}, b_{f} , c_{f}[/tex] is fixed  and the environment where [tex]b_{f} and c _{f}[/tex] is the same hence a peak is formed and another peak is formed by [tex]a_{f}[/tex] that makes the number of peaks = 2 peaks

High temperature

In high temperature [tex]a_{f}, b_{f} , c_{f}[/tex]  exchange very fast therefore one peak is formed for all, since the fast exchanges makes NMR machine to take an average and produce just one peak for all

Answer:

Velocity of 5 cm diameter pipe is 1.38 m/s

Explanation:

Use following equation of Relation between the Reynolds numbers of both pipes

[tex]Re_{5}[/tex] = [tex]Re_{12}[/tex]

[tex]\sqrt{\frac{V_{5}XD_{5} }{v_{5}}}[/tex]= [tex]\sqrt{\frac{V_{12}XD_{12} }{v_{12}}}[/tex]

[tex]Re_{5}[/tex] = Reynold number of water pipe

[tex]Re_{12}[/tex] = Reynold number of oil pipe

[tex]V_{5}[/tex] = Velocity of water 5 diameter pipe = ?

[tex]V_{12}[/tex] = Velocity of oil 12 diameter pipe = 2.30

[tex]v_{5}[/tex] = Kinetic Viscosity of water = 1 x [tex]10^{-6}[/tex] [tex]m^{2}[/tex]/s

[tex]v_{12}[/tex] = Kinetic Viscosity of oil =  4 x [tex]10^{-6}[/tex] [tex]m^{2}[/tex]/s

[tex]D_{5}[/tex] = Diameter of pipe used for water = 0.05 m

[tex]D_{12}[/tex] = Diameter of pipe used for oil = 0.12 m

Use the formula

[tex]\sqrt{\frac{V_{5}XD_{5} }{v_{5}}}[/tex]= [tex]\sqrt{\frac{V_{12}XD_{12} }{v_{12}}}[/tex]

By Removing square rots on both sides

[tex]{\frac{V_{5}XD_{5} }{v_{5}}}[/tex]= [tex]{\frac{V_{12}XD_{12} }{v_{12}}}[/tex]

[tex]{V_{5}[/tex]= [tex]{\frac{V_{12}XD_{12} }{v_{12}XD_{5}\\}}[/tex]x[tex]v_{5}[/tex]

[tex]{V_{5}[/tex]= [ (0.23 x 0.12m ) / (4 x [tex]10^{-6}[/tex] [tex]m^{2}[/tex]/s) x 0.05 ] 1 x [tex]10^{-6}[/tex] [tex]m^{2}[/tex]/s

[tex]{V_{5}[/tex] = 1.38 m/s

This question is incomplete, the complete question is;

Find the magnitude of the steady-state response of the system whose system model is given by

dx(t)/dt + x(t) = f(t)

where f(t) = 2cos8t.  Keep 3 significant figures

Answer: The steady state output x(t) = 0.2481 cos( 8t - 45° )

Explanation:

Given that;

dx(t)/dt + x(t) = f(t)  where f(t) = 2cos8t

dx(t)/dt + x(t) = f(t)

we apply Laplace transformation on both sides

SX(s) + x(s) = f(s)

(S + 1)x(s) = f(s)

f(s) / x(s) = S + 1

x(s) / f(s) = 1 / (S + 1)

Therefore

transfer function = H(s) = x(s)/f(s) = 1/(S+1)

f(t) = 2cos8t →   [ 1 / ( S + 1 ) ]   →  x(t) = Acos(8t - ∅ )

A = Magnitude of steady state output

S = jw

S = j8

so

A = 2 × 1 / √( 8² + 1 ) = 2 / √ (64 + 1 )

A = 2/√65 = 0.2481

∅ = tan⁻¹( 1/1) = 45°

therefore The steady state output x(t) = 0.2481 cos( 8t - 45° )

Explanation:

cooking of duck will cost 48000

by the help of the method of rate × A + €

Answer:

[tex]10.8\ \text{lb/ft^2}[/tex]

[tex]101.96\ \text{lb/ft}^2[/tex]

Explanation:

[tex]v_1[/tex] = Velocity of car = 65 mph = [tex]65\times \dfrac{5280}{3600}=95.33\ \text{ft/s}[/tex]

[tex]\rho[/tex] = Density of air = [tex]0.00237\ \text{slug/ft}^3[/tex]

[tex]v_2=0[/tex]

[tex]P_1=0[/tex]

[tex]h_1=h_2[/tex]

From Bernoulli's law we have

[tex]P_1+\dfrac{1}{2}\rho v_1^2+h_1=P_2+\dfrac{1}{2}\rho v_2^2+h_2\\\Rightarrow P_2=\dfrac{1}{2}\rho v_1^2\\\Rightarrow P_2=\dfrac{1}{2}\times 0.00237\times 95.33^2\\\Rightarrow P_2=10.8\ \text{lb/ft^2}[/tex]

The maximum pressure on the girl's hand is [tex]10.8\ \text{lb/ft^2}[/tex]

Now [tex]v_1[/tex] = 200 mph = [tex]200\times \dfrac{5280}{3600}=293.33\ \text{ft/s}[/tex]

[tex]P_2=\dfrac{1}{2}\rho v_1^2\\\Rightarrow P_2=\dfrac{1}{2}\times 0.00237\times 293.33^2\\\Rightarrow P_2=101.96\ \text{lb/ft}^2[/tex]

The maximum pressure on the girl's hand is [tex]101.96\ \text{lb/ft}^2[/tex]

Answer:

Yes it is possible.

Explanation:

This problem is about to possibility to have alloy of iron-carbon for which mass fraction of ferrite, [tex]$W_{\alpha} = 0.846$[/tex]  and proeutectoid cementite, [tex]$W_{Fe_3C}=0.049$[/tex]

An alloy formation is possible when the composition values of the two alloy are equal.

Now writing the expression for the mass fraction of total ferrite, we have

[tex]$W_{\alpha}=\frac{C_{Fe_3C}-C_0}{C_{Fe_3C}-C_{\alpha}}$[/tex]

[tex]$0.846}=\frac{6.70-C_0}{6.70-0.022}$[/tex]

[tex]$5.649588 = 6.70 - C_0$[/tex]

[tex]$\therefore C_0 = 1.05 $[/tex] wt. % of C

Now write the expression for the mass fraction of the proeutectoid cementite :

[tex]$W_{Fe_3C}=\frac{C_1-0.76}{5.94}$[/tex]

[tex]$0.049=\frac{C_1-0.76}{5.94}$[/tex]

[tex]$C_1 = 1.05$[/tex] % wt. C

Since, [tex]$C_0 =C_1$[/tex], it is possible to have an alloy of iron - carbon.

Answer:

the closest distance the center of gravity can be behind the front axle to have the vehicle achieve its maximum acceleration from rest on good, wet pavement is 47.8 in

Explanation:

Given that;

Weight of car W = 2600 lb

power = 80 hp = 44000 lb ft/s

Engine rpm = 3296

gear reduction ratio e = 10

drivetrain efficiency n = 95% = 0.95

wheel radius R = 16 in  = 1.3333 ft

Length of wheel base L = 95 in =

coefficient of road adhesion u = 0.60

height of center of gravity above pavement  h = 22 in

we know that;

Coefficient of rolling resistance frl = 0.01 for good wet pavement

distance of center of gravity behind the front axle lf = ?

Maximum tractive effort (Fmax) =  (uW / L) (lf - frl h) / (1 - uh / L)

First we calculate our Fmax to help us find lf

Power = Torque × 2π × Engine rpm / 60 )

44000 = Torque ( 2π×3296 / 60)

Torque = 127.5 lb ft  

so

Fmax = Torque × e × n / R

so we substitute in our values

Fmax = 127.5 × 10 × 0.95 / 1.333

Fmax = 908.66 lb

Now we input all our values into the initial formula

(Fmax) =  (uW / L) (lf - frl h) / (1 - uh / L)

908.66 =  [(0.6×2600/95) (lf - 0.01×22)] / [1 - 0.6×22) / 95]

908.66 = (16.42( lf - 0.22)) / 0.86

781.4476 = (16.42( lf - 0.22))

47.59 = lf - 0.22

lf = 47.59 + 0.22

lf = 47.8 in

Therefore the closest distance the center of gravity can be behind the front axle to have the vehicle achieve its maximum acceleration from rest on good, wet pavement is 47.8 in

Answer:

a) volume flow rate of air at the inlet is 0.0471 m³/s

b) the velocity of the air at the exit is  8.517 m/s

Explanation:

Given that;

The electrical power Input W_elec = -1400 W = -1.4 kW

Inlet temperature of air T_in = 22°C

Inlet pressure of air p_in = 100 kPa

Exit temperature T_out = 47°C

Exit area of the dyer is A_out = 60 cm²= 0.006 m²

cp = 1.007 kJ/kg·K

R = 0.287 kPa·m3/kg·K

Using mass balance

m_in = m_out = m_air

W _elec = m_air ( h_in - h_out)

we know that h = CpT

so

W _elec = m_air.Cp ( T_in - T_out)

we substitute

-1.4 = m_air.1.007 ( 22 - 47 )

-1.4 =  - m_air.25.175

m_air = -1.4 / - 25.175

m_ air = 0.0556 kg/s

a) volume flow rate of air at the inlet

we know that

m_air = P_in × V_in

now from the ideal gas equation

P_in = p_in / RT_in

we substitute our values

= (100×10³) / ((0.287×10³)(22+273))

= 100000 / 84665

P_in = 1.18 kg/m³

therefore inlet volume flowrate will be;

V_in = m_air / P_in

= 0.0556 / 1.18

= 0.0471 m³/s

the volume flow rate of air at the inlet is 0.0471 m³/s

b) velocity of the air at the exit

the mass flow rate remains unchanged across the duct

m_ air = P_in.A_in.V_in = P_out.A_out.V_out

still from the ideal gas equation

P_out = p_out/ RT_out   ( assume p_in = p_out)

P_out = (100×10³) / ((0.287×10³)(47+273))

P_out  = 1.088 kg/m³

so the exit velocity will be;

V_out = m_air / P_out.A_out

we substitute our values

V_out = 0.0556 / ( 1.088 × 0.006)

= 0.0556 / 0.006528

= 8.517 m/s

 Therefore the velocity of the air at the exit is  8.517 m/s

Answer:

2.44 mV

Explanation:

This question has to be one of analog quantization size questions and as such, we use the formula

Q = (V₂ - V₁) / 2^n

Where

n = 12

V₂ = higher voltage, 5 V

V₁ = lower voltage, -5 V

Q = is the change in voltage were looking for

On applying the formula and substitutiting the values we have

Q = (5 - -5) / 2^12

Q = 10 / 4096

Q = 0.00244 V, or we say, 2.44 mV

Solution:

Given that :

Volume flow is, [tex]$Q_1 = 1000 \ mm^3/s$[/tex]

So, [tex]$Q_2= \frac{1000}{100}=10 \ mm^3/s$[/tex]

Therefore, the equation of a single straight vessel is given by

[tex]$F_{f_1}=\frac{8flQ_1^2}{\pi^2gd_1^5}$[/tex]    ......................(i)

So there are 100 similar parallel pipes of the same cross section. Therefore, the equation for the area is

[tex]$\frac{\pi d_1^2}{4}=1000 \times\frac{\pi d_2^2}{4} $[/tex]

or [tex]$d_1=10 \ d_2$[/tex]

Now for parallel pipes

[tex]$H_{f_2}= (H_{f_2})_1= (H_{f_2})_2= .... = = (H_{f_2})_{10}=\frac{8flQ_2^2}{\pi^2 gd_2^5}$[/tex]  ...........(ii)

Solving the equations (i) and (ii),

[tex]$\frac{H_{f_1}}{H_{f_2}}=\frac{\frac{8flQ_1^2}{\pi^2 gd_1^5}}{\frac{8flQ_2^2}{\pi^2 gd_2^5}}$[/tex]

       [tex]$=\frac{Q_1^2}{Q_2^2}\times \frac{d_2^5}{d_1^5}$[/tex]

       [tex]$=\frac{(1000)^2}{(10)^2}\times \frac{d_2^5}{(10d_2)^5}$[/tex]

       [tex]$=\frac{10^6}{10^7}$[/tex]

Therefore,

[tex]$\frac{H_{f_1}}{H_{f_2}}=\frac{1}{10}$[/tex]

or [tex]$H_{f_2}=10 \ H_{f_1}$[/tex]

Thus the answer is option A). 10

Answer is given below:

Explanation:

Answer:

Water enters a centrifugal pump axially at atmospheric pressure at a rate of 0.12 m3/s and at a velocity of 7 m/s, and leaves in the normal direction along the pump casing, as shown in Fig. PI3-39. Determine the force acting on the shaft (which is also the force acting on the bearing of the shaft) in the axial direction.

Step-by-step solution:

Step 1 of 5

Given data:-

The velocity of water is .

The water flow rate is.

Answer:

hello your question has a missing part below is the missing part

Consider the string length equal to [tex]\pi[/tex]

answer : 2cos(2t) sin(2x) - 10cos(10t)sin(10x)

Explanation:

Given string length = [tex]\pi[/tex]

distorted function f(x) = 2sin(2x) - 10sin(10x)

Determine the wave formed in the string

attached below is a detailed solution of the problem

Answer:

Option B (Cold working) would be the correct alternative.

Explanation:

The other possibility isn't linked to the given scenario. Therefore the alternative above is the right one.

dutile is the correct answer

Answer:

C. The meter will display a negative sign.

Explanation:

If you use an analog voltmeter and you measure voltage with reverse polarity you will damage it. But in this case we are using a digital multimeter. This kind of multimeter is designed to be able to deal with positive and negative voltages

Solution :

Sieve Size (in)                   Weight retain(g)

3                                         1.62

2                                         2.17

[tex]$1\frac{1}{2}$[/tex]                                       3.62

[tex]$\frac{3}{4}$[/tex]                                        2.27

[tex]$\frac{3}{8}$[/tex]                                        1.38

PAN                                    0.21

Given :

Sieve       weight       % wt. retain    % cumulative       % finer

size        retained                               wt. retain

No. 4        59.5            10.225%          10.225%            89.775%

No. 8        86.5            14.865%          25.090%           74.91%

No. 16       138              23.7154%        48.8054%         51.2%

No. 30      127.8           21.91%              70.7154%          29.2850%

No. 50      97               16.6695%         87.3849%         12.62%

No. 100     66.8            11.4796%         98.92%              1.08%

Pan            6.3               1.08%              100%                   0%

                581.9 gram

Effective size = percentage finer 10% ([tex]$$D_{20}[/tex])

0.149 mm, N 100, % finer 1.08

0.297, N 50 , % finer 12.62%

x  ,   10%

[tex]$y-1.08 = \frac{12.62 - 1.08}{0.297 - 0.149}(x-0.149)$[/tex]

[tex]$(10-1.08) \times \frac{0.297 - 0.149}{12.62 - 1.08}+ 0.149=x$[/tex]

x = 0.2634 mm

Effective size, [tex]$D_{10} = 0.2643 \ mm$[/tex]

Now, N 16 (1.19 mm)  ,  51.2%

N 8 (2.38 mm)  ,  74.91%

x,  60%

[tex]$60-51.2 = \frac{74.91-51.2}{2.38-1.19}(x-1.19)$[/tex]

x = 1.6317 mm

[tex]$\therefore D_{60} = 1.6317 \ mm$[/tex]

Uniformity co-efficient = [tex]$\frac{D_{60}}{D_{10}}$[/tex]

   [tex]$Cu= \frac{1.6317}{0.2643}$[/tex]

Cu = 6.17

Now, fineness modulus = [tex]$\frac{\Sigma \text{\ cumulative retain on all sieve }}{100}$[/tex]

[tex]$=\frac{\Sigma (10.225+25.09+48.8054+70.7165+87.39+98.92+100)}{100}$[/tex]

= 4.41

which lies between No. 4  and No. 5 sieve [4.76 to 4.00]

So, fineness modulus = 4.38 mm