A cylinder 10 mm in diameter is pulled with a stress of 150 MPa. The diameter elastically decreased by 0.007 mm. Determine Poisson's ratio if the material has a elastic modulus of 100 GPa.

Answer:

a) The stresses due to bending moments is much more than the stresses from transverse shear.

c) The deformations due to bending moments is much more than the deformations from transverse shear.

Explanation:

Strain in an object suspended is a function of the stress which the suspended body passed through. The stress which is the function of the force experienced by the body over a given area helps is straining the moment. This lead to the strain energy from bending moment being greater than the strain energy from a transverse shear force.

Answer:

b

Explanation:

Answer:

± 0.003 ft

Explanation:

Since our distance is 10,000 ft and we need to use a full tape measure of 100 ft. We find that 10,000 = 100 × 100.

Let L' = our distance and L = our tape measure

So, L' = 100L

Now by error determination ΔL' = 100ΔL

Now ΔL' = ± 0.30 ft

ΔL = ΔL'/100

= ± 0.30 ft/100

= ± 0.003 ft

So, the maxim error per tape is ± 0.003 ft

Answer:

115200 and  460800

Explanation:

which of the above listed Baud rate can you choose from

Given Baud rate : 2400,4800,9600,19200,38400, 115200, 460800

The Total bytes = 98 data bytes + 2 extra label bytes for every 10 ms

                           = 100 bytes for every 10 ms

hence the data rate per second

= [tex]\frac{100 * 8}{10*10^{-3} }[/tex]  = 80000

minimum required Baud rate = 80000

Therefore The Baud rate that can be chosen from are :  115200 and  460800

Answer:

Method B is the more efficient way of heating the water.

Explanation:

Method B is more efficient because by placing a heating element in the water as in described in method B, the heat that is lost to the surroundings is minimized which implies that more heat is supplied directly to the water. Therefore, more heating is achieved with a lesser amount of electrical energy input. Whereas placing the pan on a range means more heat losses to the surrounding and as such it will take a longer time for the water to heat up and also take more electrical energy.

Answer:

The total mass in the tank = 0.45524  kg

The amount of heat transferred = 3426.33 kJ

Explanation:

Given that:

The volume of the tank V = 0.06 m³

The pressure of the liquid and the vapor of H2O (p) = 15 bar

The initial quality of the mixture [tex]\mathbf{x_{initial} - 0.20}[/tex]

By applying the energy rate balance equation;

[tex]\dfrac{dU}{dt} = Q_{CV} - m_eh_e[/tex]

where;

[tex]m_e =- \dfrac{dm_{CV}}{dt}[/tex]

Thus, [tex]\dfrac{dU}{dt} =Q_{CV} + \dfrac{dm_{CV}}{dt}h_e[/tex]

If we integrate both sides; we have:

[tex]\Delta u_{CV} = Q_{CV} + h _e \int \limits ^2_1 \ dm_{CV}[/tex]

[tex]m_2u_2 - m_1 u_1 = Q_{CV} + h_e (M_2-m_1) \ \ \ --- (1)[/tex]

We obtain the following data from the saturated water pressure tables, at p = 15 bar.

Since:

[tex]h_e =h_g[/tex]

Then: [tex]h_g = h_e = 2792.2 \ kJ/kg[/tex]

[tex]v_f = 1.1539 \times 10^{-3} \ m^3 /kg[/tex]

[tex]v_g = 0.1318 \ m^3/kg[/tex]

Hence;

[tex]v_1 = v_f + x_{initial} ( v_g-v_f)[/tex]

[tex]v_1 = 1.1529 \times 10^{-3} + 0.2 ( 0.1318-1.159\times 10^{-3} )[/tex]

[tex]v_1 = 0.02728 \ m^3/kg[/tex]

Similarly; we obtained the data for [tex]u_f \ \& \ u_g[/tex] from water pressure tables at p = 15 bar

[tex]u_f = 843.16 \ kJ/kg\\\\ u_g = 2594.5 \ kJ/kg[/tex]

Hence;

[tex]u_1 = u_f + x_{initial } (u_g -u_f)[/tex]

[tex]u_1 =843.16 + 0.2 (2594.5 -843.16)[/tex]

[tex]u_1 = 1193.428[/tex]

However; the initial mass [tex]m_1[/tex] can be calculated by using the formula:

[tex]m_1 = \dfrac{V}{v_1}[/tex]

[tex]m_1 = \dfrac{0.06}{0.02728}[/tex]

[tex]m_1 = 2.1994 \ kg[/tex]

From the question, given that the final quality; [tex]x_2 = 1[/tex]

[tex]v_2 = v_f + x_{final } (v_g - v_f)[/tex]

[tex]v_2 = 1.1539 \times 10^{-3} + 1(0.1318 -1.1539 \times 10^{-3})[/tex]

[tex]v_2 = 0.1318 \ m^3/kg[/tex]

Also;

[tex]u_2 = u_f + x_{final} (u_g - u_f)[/tex]

[tex]u_2 = 843.16 + 1 (2594.5 - 843.16)[/tex]

[tex]u_2 = 2594.5 \ kJ/kg[/tex]

Then the final mass can be calculated by using the formula:

[tex]m_2 = \dfrac{V}{v_2}[/tex]

[tex]m_2 = \dfrac{0.06}{0.1318}[/tex]

[tex]m_2 = 0.45524 \ kg[/tex]

Thus; the total mass in the tank = 0.45524  kg

FInally; from the previous equation (1) above:

[tex]m_2u_2 - m_1 u_1 = Q_{CV} + h_e (M_2-m_1) \ \ \ --- (1)[/tex]

[tex]Q = (m_2u_2-m_1u_1) - h_e(m_2-m_1)[/tex]

Q = [(0.45524)(2594.5) -(2.1994)(1193.428)-(2792.2)(0.45524-2.1994)]

Q = [ 1181.12018 - 2624.825543 - (2792.2)(-1.74416 )]

Q = 3426.33 kJ

Thus, the amount of heat transferred = 3426.33 kJ

Answer:

mountain ranges may be

Answer:

a) a1 : This is the incident voltage at port 1

b) b1 : This is the deflected voltage at port 1 ;

      b1 = [tex]S_{21} a_{1} + S_{22} a_{2}[/tex]

c) S11 ; This is the input port voltage reflection coefficient when the input voltage is at port 1

S11 = [tex]\frac{V1^-}{V1^+} |v2^+=0[/tex]

d) S12 : this is the gross voltage gain

S12 = [tex]\frac{V1^-}{V2^+}| v1 ^+[/tex]

e) S21 : This is the forward voltage gain

    S21 = [tex]\frac{V2^-}{V1^+} | v2^+[/tex]

f) S22 : output port voltage reflection coefficient

   S22 = [tex]\frac{v2^-}{v2^+} | v1^+ = 0[/tex][tex]\frac{v2^-}{v2^+} | v1^+ = 0[/tex]

Explanation:

a) a1 : This is the incident voltage at port 1

b) b1 : This is the deflected voltage at port 1 ;

      b1 = [tex]S_{21} a_{1} + S_{22} a_{2}[/tex]

c) S11 ; This is the input port voltage reflection coefficient when the input voltage is at port 1

S11 = [tex]\frac{V1^-}{V1^+} |v2^+=0[/tex]

d) S12 : this is the gross voltage gain

S12 = [tex]\frac{V1^-}{V2^+}| v1 ^+[/tex]

e) S21 : This is the forward voltage gain

    S21 = [tex]\frac{V2^-}{V1^+} | v2^+[/tex]

f) S22 : output port voltage reflection coefficient

   S22 = [tex]\frac{v2^-}{v2^+} | v1^+ = 0[/tex][tex]\frac{v2^-}{v2^+} | v1^+ = 0[/tex]

Answer:

hello your question is incomplete attached below is the missing part of the  question

Consider an inverter operating a power supply voltage VDD. Assume that matched condition for this inverter. Make the necessary assumptions to get to an answer for the following questions.

answer : Nd ∝ rt

Explanation:

Determine how the delay and active power per device will change as the doping density of N- and P-MOSFET increases

Pactive ( active power ) = Efs * F

Pactive = [tex]\frac{q^2Nd^2*Xn^2}{6Eo} * f[/tex]

also note that ; Pactive ∝ Nd2 (

tD = K . [tex]\frac{Vdd}{(Vdd - Vt )^2}[/tex]  since K = constant

Hence : Nd ∝ rt

Answer:

✔️a healthy mind resides in a healthy body.

Explanation:

The seers were of the opinion that "a healthy mind resides in a healthy body."

Just like the English translation of a famous quotation from Thales, pre-Socratic Greek philosopher puts it "a sound mind in a sound body"; which tries to demonstrate the close connections that exists in bodily well-being and one's ability to enjoy life.

The seers were actually of the opinion that a healthy mind resides in a healthy body. It implies that there is connection between the body and the mind. When the body catches an illness, the mind and other parts of the body are affected. When our minds are not healthy, it affects the effective functioning of the body.

So, a healthy mind will definitely be found in a healthy body.

✔️a healthy mind resides in a healthy body.

Explanation:

The seers were of the opinion that "a healthy mind resides in a healthy body."

Just like the English translation of a famous quotation from Thales, pre-Socratic Greek philosopher puts it "a sound mind in a sound body"; which tries to demonstrate the close connections that exists in bodily well-being and one's ability to enjoy life.

The seers were actually of the opinion that a healthy mind resides in a healthy body. It implies that there is connection between the body and the mind. When the body catches an illness, the mind and other parts of the body are affected. When our minds are not healthy, it affects the effective functioning of the body.

So, a healthy mind will definitely be found in a healthy body.

Answer:

This graph shows linear

y = f(x) and y = g(x).

Find the solution to the equation f(x) - g(x) = 0

Answer:

74,4 litros

Explanation:

Dado que

W = nRT ln (Vf / Vi)

W = 3000J

R = 8,314 JK-1mol-1

T = 58 + 273 = 331 K

Vf = desconocido

Vi = 25 L

W / nRT = ln (Vf / Vi)

W / nRT = 2.303 log (Vf / Vi)

W / nRT * 1 / 2.303 = log (Vf / Vi)

Vf / Vi = Antilog (W / nRT * 1 / 2.303)

Vf = Antilog (W / nRT * 1 / 2.303) * Vi

Vf = Antilog (3000/1 * 8,314 * 331 * 1 / 2,303) * 25

Vf = 74,4 litros

Answer:

i) r_t = 0.5101 m

ii) m' = 106.73 kg/s

iii) R_s = 1.26

P = 2359.8 kW

iv) β2 = 55.63°

Explanation:

We are given;

Stagnation pressure; T_01 = 290 K

Inlet velocity; C1 = 145 m/s

Cp for air = 1005 kJ/(kg·K)

Mach number; M = 0.96

Ratio of specific heats; γ = 1.4

Stagnation pressure; P_01 = 1 bar

rotational speed; N = 5500 rpm

Work done factor; τ = 0.92

Isentropic effjciency; η = 0.9

Stagnation temperature rise; ΔT_s = 22 K

i) Formula for Stagnation temperature is given as;

T_01 = T1 + C1/(2Cp)

Thus,making T1 the subject, we havw;

T1 = T_01 - C1/(2Cp)

Plugging in the relevant values, we have;

T1 = 290 - (145/(2 × 1005))

T1 = 289.93 K

Formula for the mach number relative to the tip is given by;

M = V1/√(γRT1)

Where V1 is relative velocity at the tip and R is a gas constant with a value of 287 J/Kg.K

Thus;

V1 = M√(γRT1)

V1 = 0.96√(1.4 × 287 × 289.93)

V1 = 0.96 × 341.312

V1 = 327.66 m/s

Now, tip speed is gotten from the velocity triangle in the image attached by the formula;

U_t = √(V1² - C1²)

U_t = √(327.66² - 145²)

U_t = √86336.0756

U_t = 293.83 m/s

Now relationship between tip speed and tip radius is given by;

U_t = (2πN/60)r_t

Where r_t is tip radius.

Thus;

r_t = (60 × U_t)/(2πN)

r_t = (60 × 293.83)/(2π × 5500)

r_t = 0.5101 m

ii) Now mean radius from derivations is; r_m = 1.5h

While relationship between mean radius and tip radius is;

r_m = r_t - h/2

Thus;

1.5h = 0.5101 - 0.5h

1.5h + 0.5h = 0.5101

2h = 0.5101

h = 0.5101/2

h = 0.2551

So, r_m = 1.5 × 0.2551

r_m = 0.3827 m

Formula for the area is;

A = 2πr_m × h

A = 2π × 0.3827 × 0.2551

A = 0.6134 m²

Isentropic relationship between pressure and temperature gives;

P1 = P_01(T1/T_01)^(γ/(γ - 1))

P1 = 1(289.93/290)^(1.4/(1.4 - 1))

P1 = 0.9992 bar = 0.9992 × 10^(5) N/m²

Formula for density is;

ρ1 = P1/(RT1)

ρ1 = 0.9992 × 10^(5)/(287 × 289.93)

ρ1 = 1.2 kg/m³

Mass flow rate at compressor inlet is;

m' = ρ1 × A × C1

m' = 1.2 × 0.6134 × 145

m' = 106.73 kg/s

iii) stagnation pressure ratio is given as;

R_s = (1 + ηΔT_s/T_01)^(γ/(γ - 1))

R_s = (1 + (0.9 × 22/290))^(1.4/(1.4 - 1))

R_s = 1.26

Work is;

W = C_p × ΔT_s

W = 1005 × 22

W = 22110 J/Kg

Power is;

P = W × m'

P = 22110 × 106.73

P = 2359800.3 W

P = 2359.8 kW

iv) We want to find the rotor angle.

now;

Tan β1 = U_t/C1

tan β1 = 293.83/145

tan β1 = 2.0264

β1 = tan^(-1) 2.0264

β1 = 63.73°

Formula for Stagnation pressure rise is given by;

ΔT_s = (τ•U_t•C1/C_p) × tan(β1 - β2)

Plugging in the relevant values;

22 = (0.92 × 293.83 × 145/1005) × (tan 63.73 - tan β2)

(tan 63.73 - tan β2) = 0.5641

2.0264 - 0.5641 = tan β2

tan β2 = 1.4623

β2 = tan^(-1) 1.4623

β2 = 55.63°

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:

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.

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;

A thin, flat pate that is 0.2m by 0.2m on a side is oriented parallel to an atmospheric airstream having a velocity of 40m/s. The air is at a temperature of T∞ = 20 °C, while the plate is maintained at Ts = 120°C. The air flows over the top and bottom surfaces of the plate, and measurement of the drag force reveals a value of 0.075 N.

What is the rate of heat transfer from both sides of the plate to the air?

Answer:

the rate of heat transfer from both sides of the plate to the air is 236.54 W

Explanation:

Given the data in the question,

first we calculate the  Reynold's number for the flow

Re = pu∞d / Ц

Re = (1.12 × 40 × 0.2) / 1.983 × 10⁻⁵

Re = 451840

Now the Local skin friction coefficient is given as;

Cfx =  T / ( 1/2pu∞²)

Cfx = (Fd/A) / ( 1/2pu∞²)

Cfx = (0.075/(2×0.2×0.2)) / ( 1/2 × 1.12 × 40²)

= 0.9375 / 896

= 0.0010463

Cfx = 1.0463 × 10⁻³

Apply Reynold's- cOLBURN analogy

Cfx/2 = StₓPr^2/3

so

1.0463 × 10⁻³ / 2 = (h/pu∞Cp) × ( 0.711)^2/3

5.2315 × 10⁻⁴ × 1.12 × 40 × 1.005 × 1000 = h(0.711)^2/3

h = 23.554 / 0.7966

h = 29.56 W/m².K

so

The heat transfer rate from both the sides of the plate will be;

Q = 2 × 29.56 × 0.2 × 0.2 × ( 120 - 20 )

Q = 236.54 W

Therefore the rate of heat transfer from both sides of the plate to the air is 236.54 W

Answer:

1)ΔL = 0.616 mm

2)Δd = 0.00194 mm

Explanation:

We are given;

Force; F = 52900 N

Initial length; L_o = 207 mm = 0.207 m

Diameter; d_o = 19.2 mm = 0.0192 m

Elastic modulus; E = 61.4 GPa = 61.4 × 10^(9) N/m²

Now, from Hooke's law;

E = σ/ε

Where; σ is stress = force/area = F/A

A = πd²/4 = π × 0.0192²/4

A = 0.00009216π

σ = 52900/0.00009216π

ε = ΔL/L_o

ε = ΔL/0.207

Thus,from E = σ/ε, we have;

61.4 × 10^(9) = (52900/0.00009216π) ÷ (ΔL/0.207)

Making ΔL the subject, we have;

ΔL = (52900 × 0.207)/(61.4 × 10^(9) × 0.00009216π)

ΔL = 0.616 × 10^(-3) m

ΔL = 0.616 mm

B) Poisson's ratio is given as;

υ = ε_x/ε_z

ε_x = Δd/d_o

ε_z = ΔL/L_o

Thus;

υ = (Δd/d_o) ÷ (ΔL/L_o)

Making Δd the subject gives;

Δd = (υ × d_o × ΔL)/L_o

We are given Poisson's ratio to be 0.34.

Thus;

Δd = (0.34 × 19.2 × 0.616)/207

Δd = 0.00194 mm

Answer:

robotic technology    

Explanation:

Innovation is nothing but the use of various things such as ideas, products, people to build up a solution for the benefit of the human. It can be any product or any solution which is new and can solve people's problems.

Innovation solution makes use of technology to provide and dispatch new solutions or services which is a combination of both technology and ideas.

One such example of an innovative solution we can see is the use of "Robots" in medical science or in any military operations or rescue operation.

Sometimes it is difficult for humans to do everything or go to everywhere. Thus scientist and engineers have developed many advance robots or machines using new ideas and technology to find solutions to these problems.

Using innovations and technologies, one can find solutions to many problems which is difficult for the peoples. Robots can be used in any surveillance operation or in places of radioactive surrounding where there is a danger of humans to get exposed to such threats. They are also used in medical sciences to operate and support the patient.  

Answer:

A. 72.34mol/min

B. 76.0%

Explanation:

A.

We start by converting to molar flow rate. Using density and molecular weight of hexane

= 1.59L/min x 0.659g/cm³ x 1000cm³/L x 1/86.17

= 988.5/86.17

= 11.47mol/min

n1 = n2+n3

n1 = n2 + 11.47mol/min

We have a balance on hexane

n1y1C6H14 = n2y2C6H14 + n3y3C6H14

n1(0.18) = n2(0.05) + 11.47(1.00)

To get n2

(n2+11.47mol/min)0.18 = n2(0.05) + 11.47mol/min(1.00)

0.18n2 + 2.0646 = 0.05n2 + 11.47mol/min

0.18n2-0.05n2 = 11.47-2.0646

= 0.13n2 = 9.4054

n2 = 9.4054/0.13

n2 = 72.34 mol/min

This value is the flow rate of gas that is leaving the system.

B.

n1 = n2 + 11.47mol/min

72.34mol/min + 11.47mol/min

= 83.81 mol/min

Amount of hexane entering condenser

0.18(83.81)

= 15.1 mol/min

Then the percentage condensed =

11.47/15.1

= 7.59

~7.6

7.6x100

= 76.0%

Therefore the answers are a.) 72.34mol/min b.) 76.0%

Please refer to the attachment .

Answer:64.10 Btu/lbm

Explanation:

Work done in an isothermally compressed steady flow device is expressed as

Work done = P₁V₁ In { P₁/ P₂}

Work done=RT In { P₁/ P₂}

where P₁=13 psia

          P₂= 80 psia

Temperature =°F Temperature is convert to  °R

T(°R) = T(°F) + 459.67

T(°R) = 55°F+ 459.67

=514.67T(°R)

According to the properties of molar gas, gas constant and critical properties table, R  which s the gas constant of air is given as 0.06855 Btu/lbm

Work = RT In { P₁/ P₂}

0.06855 x 514.67 In { 13/ 80}

=0.06855 x 514.67 In {0.1625}

= 0.06855 x 514.67  x -1.817

=- 64.10Btu/lbm

The required work therefore for this  isothermal compression is 64.10 Btu/lbm

Answer:

design

Explanation:

Design is a process used to solve problems systematically.

Human beings have specific needs and desires, which require a design process to interpret those needs and make them real from a product or service.

Design uses specific methods and techniques integrating ideals, creativity, technology and innovation to satisfy users' needs and solve problems.

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

Answer:

10uF

Explanation:

A higher value of capacitance is the best option when we are trying to filter power supply outputs in other to reduce hum.

The greater the capacitance or the voltage of a circuit is, the more energy it can the particular circuit can store. When capacitors are being connected in series, the total value of the capacitance reduces but contrarily, the voltage of the same system increases anyway. Connecting circuits in parallel helps to keep the voltage rating the same but on the other hand, it increases the total capacitance.

A 10 μF capacitor is better.

This is because, to filter out high frequency noise, our capacitor is connected in parallel with the voltage supply. This parallel connection causes the capacitance of the circuit to increase but the voltage stays constant.

Since there is an increase in capacitance, this causes the circuit to filter out high frequency noise.

So, a high value capacitance connected in parallel with the voltage source is a better filter for high frequency noise.

So, the 10 μF capacitor is better.

Learn more about capacitors here:

https://brainly.com/question/24927491

Answer:

8 kmol/s

Explanation:

From the given information:

The combustion reaction equation for Octane in a stoichiometric condition can be expressed as:

[tex]C_{8}H_{18} +12.5(O_2 + \dfrac{79}{21} N_2) \to 9H_2O +8CO_2 + 12.5(\dfrac{79}{21}N_2)[/tex]

[tex]C_{8}H_{18} +12.5(O_2 + 3.76N_2) \to 9H_2O +8CO_2 + 12.5(3.76 \ N_2)[/tex]

In the combustor, it is said that 60% of excess air and 1 mole of Octane is present.

Thus;

the air supplied = 1.6  × 12.5 = 20

The equation can now be re-written as:

[tex]C_{8}H_{18} +20(O_2 + 3.76N_2) \to 9H_2O +8CO_2 + 7.5 \ O_2+ 75.2 \ N_2[/tex] because for 1 mole of Octane, 8 moles of CO2 can be found in the combustion product.

Thus, for 1 kmol/s of Octane also produce 8 kmol/s of CO2.

∴

The mole flow rate in Kmol/s of CO2 in the product stream = 8 kmol/s

Answer: overall composition ⇒ 87 wt% { AL₂O₃] + 13% wt { SiO₂}

Explanation:

Given that;

from the phase diagram SiO₂ - Al₂O₃

alumina at 1400°C

mullite + alumina ranges from 74 - 100% wt

so for 50% mullite and 50wt% alumina

we have;

50/100 = 100 - x /  100 - 74

0.5 = 100 - x / 26

0.5 × 26 = 100 - x

13 = 100 - x

x = 100 - 13

x = 87 wt% { AL₂O₃]

[ 100% - 87% = 13%] 13% wt SiO₂

So overall composition ⇒ 87 wt% { AL₂O₃] + 13% wt { SiO₂}

Answer:

Hello the needed data given is not properly arranged attached below is the properly arranged data

Answer:

b) 62.5 * 10^3 MPa

c) ≈ 285 MPa

d)  370Mpa

e)  16%

Explanation:

Given Data:

cylindrical aluminum diameter = 12.8 mm

Gauge length = 50.8 mm

A) plot of engineering stress vs engineering strain

attached below

B ) calculate Modulus of elasticity

Modulus of elasticity = Δб / Δ ε

                                   = ( 200 - 0 ) / (0.0032 - 0 ) = 62.5 * 10^3 MPa

C) Determine the yield strength

at strain offset = 0.002

hence yield strength ≈ 285 MPa

D) Determine tensile strength of the alloy

The tensile strength can be approximated at 370Mpa because that is where it corresponds to the maximum stress on the stress  vs strain ( complete plot )

E) Determine approximate ductility in percent elongation

ductility in percent elongation = plastic strain at fracture * 100

total strain = 0.165 , plastic strain = 0.16

therefore Ductility in percent elongation = 0.16 * 100 = 16%

Answer:

155fts

Explanation:

We apply the bernoulli's equation to get the depth of water.

We have the following information

P1 = pressure at top water surface = 0

V1 = velocity at too water surface = 0

X1 = height of water surface = h

Hf = friction loss = 0

P2 = pressure at exit = 0

V2 = velocity at exit if penstock = 100ft/s

X2 = height of penstock = 0

g = acceleration due to gravity = 32.2ft/s²

Applying these values to the equation

0 + 0 + h = 0 + v2²/2g +0 + 0

= h = 100²/2x32.2

= 10000/64.4

= 155.28ft

= 155