An inductor has an inductance of 0.025 H and a wire resistance of . How long will it take the current to reach its full Ohm’s law value?

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

Answer:

b

Explanation:

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:

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:

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:

± 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:

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

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:

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:

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. 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:

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:

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:

mountain ranges may be

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:

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

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:

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:

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:

✔️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:

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:

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: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

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

Answer:

Yes, they can all be present in the same malware because each of them perform slightly differing functions.

Explanation:

Backdoor is a software which when placed into your computer will permit hackers to easily gain reentry into your computer. This can happen even after you have already patched the flaw that they have used to hack your system before.

A bot is a program that does the same task in a continuous manner akin to when you use a blender by pressing the button to blend what you have put into it.

A keylogger is a part of a hidden software that monitors and records everything you type on your computer keyboard after which it writes it onto a file, with the hopes of capturing relevant information such as your bank account number and even passwords and other sensitive means of identification.

A Spyware is somehow similar to a keylogger just that it steals information from your computer and sends it to someone else.

A root kit is a bad software that is capable of modifying the operating system or other privileged access devices in order to gain continuous access into your system for the purpose of gathering of information and/or reducing the system’s functionality.

Yes, they can all be present in the same malware because each of them perform slightly differing functions.

Answer:

D. Both "Materials" and "Packaging"

Explanation:

Product liability may refer to the manufacturer or the seller being held responsible or liable for providing any defective product into the hands of the consumer or the customer. Responsibility or liability for a defective product which causes injuries lies with all the sellers of the product from the manufacturer to the distributor to the seller.

There are majorly three product defects. They are :

1. Manufacturing defect

2. Design defect

3. Marketing defect

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:

This graph shows linear

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

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

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