Consider a circuit element, with terminals a and b, that has vab= -12V and iab= 3A. Over a period of 2 seconds, how much charge moves through the element? If electrons carry the charge, which terminal do they enter? How much energy is transferred? Is it delivered to the element or taken from it?

Answer:

155 KJ

Explanation:

The total enthalpy is given by

ΔH=ΔU + PV

Where;

ΔH = enthalpy

ΔU = internal energy = 25000 kJ/kg/ 5 kg = 5000 KJ

P = 150 kPa = 150,000 Pa

V =  1 m3

ΔH=  5000 + (150,000 * 1)

ΔH=  155 KJ

Answer:

β = [tex]\frac{c}{\sqrt{km} }[/tex] =  0.7071 ≈ 1 ( damping condition )

closed-form expression for the response is attached below

Explanation:

Given :  x + 2x + 2x = 0   for Xo = 0 mm and Vo = 1 mm/s

computing a solution :

M = 1,

c = 2,

k = 2,

Wn = [tex]\sqrt{\frac{k}{m} }[/tex]  = [tex]\sqrt{2}[/tex]  

next we determine the damping condition using the damping formula

β = [tex]\frac{c}{\sqrt{km} }[/tex] =  0.7071 ≈ 1

from the condition above it can be said that the damping condition indicates underdamping

attached below is the closed form expression for the response

Answer:

The answer is "Option A".

Explanation:

Series:

[tex]9, 94, 916, 964, 9256, ........[/tex]

Solving the above series:

[tex]\to 9\\ \to 9(4) =94\\\to 9 (4^2) = 9(16) =916\\\to 9 (4^3) = 9(64) =964\\\to 9 (4^4) = 9(256) =9256\\\to 9 (4^5) = 9(1024) =91024\\\to 9 (4^6) = 9(4096) =94096\\[/tex]

So, the series is:  [tex]9, 94, 916, 964, 9256, 91024, 94096, .................[/tex]

Answer: the air flow rate a is 90 kg/sec; Option b) 90 kg/sec is the correct answer

Explanation:

Given that;

product of combustor flow rate m = 100 kg/s

air-fuel = 9

Airflow rate = ?

⇒We know that in the combustor, air fuel are mixed and then ignited,

⇒air fuel products are exited at the combustor

let air and fuel be a and b respectively

⇒ a + b = 100 kg/sec ----- let this be equation 1

now

⇒ air / fuel = 9

a / b = 9

a = 9b -----------let this be equation 2

now input a = 9b in equation 1

9b + b = 100 kg/sec

10b = 100 kg/sec

b = 10 kg/sec

we know that

a = 9b

so a = 9 × 10 = 90 kg/sec

Therefore the air flow rate a is 90 kg/sec

Answer:

In the case of the production Line, we know that,

No of gasoline cans = 5

probability that 1st can is out of tolerance = 0.03

probability that 2nd can is out of tolerance = 0.03

.

.

probability that the 5th can is out of tolerance = 0.03

Therefore,

probability of 1st can out of tolerance + probability of 1st can not out of tolerance = 1

Probability of 1st can not out of tolerance = 1 -- 0.03 = 0.97

probability of 2nd can not out of tolerance = 0.97

.

.

probability of 5th can not out of tolerance = 0.97

Question A:

Probability that they are all out of tolerance

= P(1st can out of tolerance) * P(2nd can out of tolerance) * P(3rd can out of tolerance) * P(4th can out of tolerance) * P(5th can out of tolerance)  

= (0.03 ) * (0.03) * (0.03) * (0.03) * (0.03) =  2.43 E⁻⁸   (2.43 ˣ 10⁻⁸)

Question B:

Probability that exactly two are out of tolerance

= P(1st can is out of tolerance) * P(2nd can is out of tolerance) * P(3rd can is not out of tolerance) * P(4th can is not out of tolerance) * P(5th can is not out of tolerance)

= (0.03) * (0.03) * (0.97) * (0.97) * (0.97) = 0.0008214057

Explanation:

Answer: attached below is the power triangles

7.13589 MVAR

Explanation:

Power ( P1 ) = 10 MW

power factor ( cos ∅ ) = 0.6 lagging

New power factor = 0.85

Calculate the reactive power of a capacitor to be connected in parallel

Cos ∅ = 0.6

therefore ∅ = 53.13°

S = P1 / cos ∅ = 16.67 MVA

Q1 = S ( sin ∅ ) = 13.33 MVAR  ( reactive power before capacitor was connected in parallel )

note : the connection of a capacitor in parallel will cause a change in power factor and reactive power while the active power will be unchanged i.e. p1 = p2

cos ∅2 = 0.85 ( new power factor )

hence ∅2 =  31.78°

Qsh ( reactive power when power factor is raised to 0.85 )

= P1 ( tan∅1 - tan∅2 )

= 10 ( 1.333 - 0.6197 )

= 7.13589 MVAR

Answer:

The hardenability increases with increasing austenite grain size, because the grain boundary area is decreasing. This means that the sites for the nucleation of ferrite and pearlite are being reduced in number, with the result that these transformations are slowed down, and the hardenability is therefore increased.

Answer:

Super critical

1.2 m

Explanation:

Q = Flow rate = [tex]12\ \text{m}^3/\text{s}[/tex]

w = Width = 3 m

d = Depth = 90 cm = 0.9 m

A = Area = wd

v = Velocity

g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

[tex]Q=Av\\\Rightarrow v=\dfrac{Q}{wd}\\\Rightarrow v=\dfrac{12}{3\times 0.9}\\\Rightarrow v=4.44\ \text{m/s}[/tex]

Froude number is given by

[tex]Fr=\dfrac{v}{\sqrt{gd}}\\\Rightarrow Fr=\dfrac{4.44}{\sqrt{9.81\times 0.9}}\\\Rightarrow F_r=1.5[/tex]

Since [tex]F_r>1[/tex] the flow is super critical.

Flow is critical when [tex]Fr=1[/tex]

Depth is given by

[tex]d=(\dfrac{Q^2}{gw^2})^{\dfrac{1}{3}}\\\Rightarrow d=(\dfrac{12^2}{9.81\times 3^2})^{\dfrac{1}{3}}\\\Rightarrow d=1.2\ \text{m}[/tex]

The depth of the channel will be 1.2 m for critical flow.

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Answer:people tend to do this when they are in a different environment they lose something or just have something going on in their life

Explanation:

Answer:

In Crystalline metals or materials, plasticity is examined from the perspective of the motion of linear defects or dislocations within the polymer chains.

Explanation:

When a temperature range below and near the glass transition temperature is reached, there is warping or contortion of structureless or malformed polymers. This warping happens as the polymer chains move over one another.

Unlike elasticity when requires or enables an object to resume its original dimensions, ductility is the quality of an element or material to change form albeit permanently.

Cheers

Answer: Option D) 298 g/mol  is the correct answer

Explanation:

Given that;

Mass of sample m = 13.7 g

pressure P = 2.01 atm

Volume V = 0.750 L

Temperature T = 399 K

Now taking a look at the ideal gas equation

PV = nRT

we solve for n

n = PV/RT

now we substitute

n = (2.01 atm x 0.750 L) / (0.0821 L-atm/mol-K x 399 K )

= 1.5075 / 32.7579

= 0.04601 mol

we know that

molar mass of the compound = mass / moles

so

Molar Mass = 13.7 g / 0.04601 mol

= 297.7 g/mol  ≈ 298 g/mol

Therefore Option D) 298 g/mol  is the correct answer

Answer:

a)  ( ∝ ) = 69.6548

b) supply power factor = 0.6709

  displacement power factor = 0.8208

Explanation:

Given data:

Nominal voltage ( E ) = 200-V

resistance (r) = 10 milliohms

constant current ( I )  = 20 amps

Phase ; 3-phase

semi-converter  with 220-v ( line-to-line ) ,  220√2  ( phase voltage )

frequency ; 60 Hz

a) determine the firing angle of thyristors

Vo = E + I*r

    = 200 + 20*10*10^-3

    = 200.2 v

attached below is the remaining part of the solution

firing angle of thyristors for charging process ( ∝ ) = 69.6548

b) determine displacement power factor and supply power factor

attached below is the detailed solution

Displacement power factor ( Dpf ) = cos ( ∝ /2 ) = 0.8208

displacement power factor = g * Dpf

                                             = 0.81747 * 0.8208 = 0.6709

Answer:

0.21

Explanation:

This would have been a fairly easy one, except for that the first part of the question is missing, and as such, I'd assume a value.

We need to use chord, so, I'm assuming the length of the chord to be 0.045 m

The Area is given by the formula

Area = span * chord

Area = 0.245 * 0.045

Area = 0.011 m²

This area gotten, is what we then divide the glider weight by to get our answer.

Lee = area / weight

Lee = 0.011 / 0.0523

Lee = 0.21

Therefore, using the values of the chord I'd assumed, the Lee of the same wing is 0.21

Answer: 0.2108

Explanation:got it correct

Answer: the magnetic flux density at the center of the loop is μ₀I  / 4πα

Explanation:

Taking a look at the diagram;

we draw an imaginary curve to complete it as a circle.

we can now apply amperes law to write;

flux density B at centre;

∫B.dl = μ₀I

now since;

∫dl = 2πα

and field direction at centre is perpendicular to the screen, so it add up , and hence constant in magnitude.

so we can be taken out of the integral ,

B( 2πα ) = μ₀I

hence ;

B_circle = B = μ₀I  / 2πα

so if we remove the half part of this;

we get a semicircle, which is what we are looking for;

Now

B_semi = 1/2.B = 1/2 × μ₀I  / 2πα

B_semi = μ₀I  / 4πα

Therefore the magnetic flux density at the center of the loop is μ₀I  / 4πα

Answer:

(a)This portion of the problem asks that we calculate, for a Pb-Mg alloy, the mass of lead in 7.5 kg of thesolidphase at 300C just below the solubility limit.From Figure 9.20, the solubility limit for thephase at

Explanation:

Answer: 1.137*10^7 Btu/h.

Explanation:

Given data:

Efficiency of the plant = 4.5percent

Net power output of the plant = 150kw

Solution:

The required collection rate

QH = W/n

= 150/0.045 * 0.94782/ 1 /60 */60 Btu/h.

= 3333.333 *3412.152Btu/h.

= 11373840 Btu/h

= 1.137*10^7 Btu/h.

Answer:

a) Normal stress :

бn =[ ( бx + бy ) / 2  + ( бx - бy ) / 2  ] cos2∅ + Txysin2∅

shear stress

Tn = ( - бx - бy ) / 2  sin2∅ + Txy cos2∅

b) principal stress :

б1 = ( бx + бy ) / 2  - [tex]\sqrt{}[/tex]( ( бx - бy ) / 2 )^2 + T^2xy

maximum shear stress:

Tmax  = ( б1 - б2) / 2 = √ (( бx - бy ) / 2 )^2 + T^2xy

Explanation:

Combined normal stress and shear stress  sketches attached below

The terms in the sketch are :

бx = tensile stress in x direction

бy =  tensile stress in y direction

Txy = y component of shear stress acting on the perpendicular plane to x axis

бn = Normal stress acting on the inclined plane EF

Tn = shear stress acting on the inclined plane EF

A) Normal and shear stresses on inclined plane

Normal stress :

бn =[ ( бx + бy ) / 2  + ( бx - бy ) / 2  ] cos2∅ + Txysin2∅

shear stress

Tn = ( - бx - бy ) / 2  sin2∅ + Txy cos2∅

B) principal and maximum shear stresses

principal stress :

б1 = ( бx + бy ) / 2  - [tex]\sqrt{}[/tex]( ( бx - бy ) / 2 )^2 + T^2xy

maximum shear stress:

Tmax  = ( б1 - б2) / 2 = √ (( бx - бy ) / 2 )^2 + T^2xy

Answer:

The answer for the question is true

Explanation:

If you get a virus or get hacked you will still have it saved

Answer:

Building technology is building technology such as coding an app or a website

Building engineering is making computers or cars or phones

Explanation:

Answer:

the motor input power is 19.42 KW

the Reactive power is 17.65 KVAR

Current is 31.56 A

Explanation:

Given that;

V = 480V

h.p = 25 hp

p.f = 0.74 lagging

n_motor = 96%

so output = 25hp

and we know that;

1hp = 746 watt

watt = hp × 1hp

so output in watt = 25 × 746 = 18650 Watt = 18.65 KW

n_motor = (output / input) × 100

96 = 1865 / Input

96Input = 1865

Input = 1865 / 96

Input = 19.42 KW

Therefore the motor input power is 19.42 KW

P = √( 3 × V × I × cos∅)

19.42 = √( 3 ×480 × I × 0.74)

I = 31.56 A

Therefore Current is 31.56 A

Q = √( 3 × V × I × sin∅)

we know that

cos∅ = 0.74

so ∅ = cos⁻¹(0.74) = 42.26

so we substitute

Q = √( 3 × 480 × 31.56 × sin(42.26))

 = 17.65 KVAR

Therefore the Reactive power is 17.65 KVAR

Answer:

The ammonia is still in the gas phase

Explanation:

Given that 1 bar is approximately = 1 atm

From;

PV=nRT

P= 310 atm

V= the unknown

n= 1

R = 0.082atm LK-1mol-1

T = 92oC + 273 = 365 K

V= nRT/P

V= 1 * 0.082 * 365/310

V = 0.0965 L = 96.5 mL

molar mass of NH3 = 17 g/mol

Molar density of NH3 = 17g/ 96.5mL = 0.176g/mL

The ammonia is still in the gas phase

Explanation:

D. B. C. A. E. Is this a good idea

Solution:

Given :

D = 6.35 cm

[tex]$\bar f = 0.005$[/tex]

[tex]$P_s = 280 \ kPa$[/tex]

[tex]$T_s= 825 K[/tex]

a). From fanno flow table (γ = 1.4)

At [tex]$M_1 = 0.12$[/tex] ,    [tex]$\left(\frac{4 \bar f L_{max}}{D}\right)_1 = 45.408$[/tex]

At [tex]$M_2 = 1$[/tex] ,      [tex]$\left(\frac{4 \bar f L_{max}}{D}\right)_2 = 0$[/tex]

∴  [tex]$\left(\frac{4 \bar f L_{max}}{D}\right)_1 - \left(\frac{4 \bar f L_{max}}{D}\right)_2 = \frac{4 \bar f L}{D}$[/tex]

[tex]$45.408 - 0 = \frac{4 \times 0.005 \times L}{0.0635}$[/tex]

[tex]$45.408 = \frac{4 \times 0.005 \times L}{0.0635}$[/tex]

L = 144.17 m

b). If L = 25 m

[tex]$\frac{4 \bar f L}{D}=\frac{4 \times 0.005 \times 25}{0.0635} = 7.874$[/tex]

From fanno flow , (γ = 1.4)

[tex]$At, M_2 = 0.26 , \frac{4 \bar f L}{D} = 7.874$[/tex]

[tex]$\frac{P_s}{P_1}=\frac{T_s}{T_1}^{\frac{\gamma}{\gamma - 1}} = (1+\frac{\gamma-1}{2}M_1^2)^{\frac{\gamma}{\gamma - 1}}$[/tex]

[tex]$\frac{280}{P_1}=\frac{825}{T_1}^{\frac{1.4}{1.4 - 1}} = (1+\frac{1.4-1}{2}(0.12)^2)^{\frac{1.4}{1.4 - 1}}$[/tex]

[tex]$\frac{280}{P_1}=\left(\frac{825}{T_1}\right)^{3.5} =1.0101$[/tex]

[tex]$P_1 = 277.2 \ kPa$[/tex]

[tex]$T_1=822.63 \ K$[/tex]

[tex]$\frac{T_2}{T_1}=\frac{1+\frac{\gamma -1}{2}M_1^2}{1+\frac{\gamma -1}{2}M_2^2}$[/tex]

[tex]$\frac{T_2}{822.63}=\frac{1+\frac{1.4 -1}{2}(0.12)^2}{1+\frac{1.4 -1}{2}(0.26)^2}$[/tex]

[tex]$\frac{T_2}{822.63}=\frac{1.00288}{1.01352}$[/tex]

[tex]$T_2=814 \ K$[/tex]

[tex]$\frac{P_2}{P_1}=\frac{M_1}{M_2}\left(\frac{1+\frac{\gamma -1}{2}M_1^2}{1+\frac{\gamma -1}{2}M_2^2}\right)^{1/2}$[/tex]

[tex]$\frac{P_2}{P_1}=\frac{0.12}{0.26}\left(\frac{1+\frac{1.4 -1}{2}(0.12)^2}{1+\frac{1.4 -1}{2}(0.26)^2}\right)^{1/2}$[/tex]

[tex]$\frac{P_2}{277.2}=0.459$[/tex]

[tex]$P_2=127.27 \ kPa$[/tex]

Answer:

Following are the solution to this question:

Explanation:

To copy 3.0 bits in 50 dollars or run at 50 dollars, it takes just 3.0 bits as well as other bits but masks, and 50 dollars.  

Instead of shifting the $ 50 by 3 bits to 6...3 bits of [tex]\$ \ 50, \$ \ 50,0*0000 0003,[/tex] This procedure instead took place at $53 and $50  

AND [tex]\$ \ 50,\$ \ 50,0*0000 000f[/tex], take 3..0  bits

SLL [tex]\$ \ 50, \$ \ 50,0*0000 0003,[/tex]Shifts the bits to 6..3

O R [tex]\$ \ 53,\$ \ 53,\$ \ 50 ,[/tex] coping to [tex]\$ \ 53[/tex]

Answer:

269.1J

Explanation:

m = 12g

c = 0.897J/g°C

∆T = 45 - 20 = 25°C

H = mc∆T = 12 × 0.897 × 27 = 269.1J

Answer:

Explanation:

Since we are considering electron and hole drift velocities, then electric field E will have to be taken into consideration as well.

Where E = V/d...... 1

Drift velocity (u) = -μE. For electron.... 2

Drift velocity (v) = μE. For hole...... 3

Given that : V = 5V and d = 10 um (micro meter)

From equation 1

E = V/d

E = 5V/10×10^-4cm

E = 5V ÷1/1000

E = 5×1000

E = 5000v/cm

From equation 2

Un = -μE.

Un = - 1350cm^2/vs × 5000

= -6750000cm/s

From equation 3

Vp = μE

= 480cm^2/vs × 5000

= 2400000cm/s

Since it was stated in the question that we should contrast between hole drift and electron drift.

6750000/2400000

= 2.8125

Hence the electron drift velocity is 2.8 times that of hole drift velocity indicating that the speed of the electron through the silicon was faster.

Answer:

the answer would be d its d

Answer:

Pretty sure the answer is "C"

Explanation:

"The media doesn't affect me at all because I'm smart enough to know the difference between right and wrong."

Answer:Total Voltage = 14V

Explanation: it is possible that a circuit  can contain more than one source of electromotive force which can cause flow of current in the same or opposite direction . When the  connection to  voltage sources  allows for current  from the voltage sources to flow in  same direction,it is termed  Series aiding  Thus, the  Total/effective voltage in a series aiding circuit is  computed as the sum of series aiding voltages .

 Here we have the series aiding voltages to be 5V and 9V ,

therefore,

Total Voltage = 5V + 9V

= 14V