Answer:
a) Real Power (kW) = 1.5 kW
b) Reactive Power (kvar) is 1.1663 KVAR
c) Apparent Power (kVA) is 1.9 KVA
d) the Power Factor cos∅ is 0.7894
e) the impedance Z in polar and rectangular form is 76 ∠ 37.87° Ω
Explanation:
Given that;
V = 380v
i = 5A
P = 1500 W
determine;
a) Real Power (kW)
P = 1500W = 1.5 kW
therefore Real Power (kW) = 1.5 kW
b) Reactive Power (kvar)
p = V×i×cos∅
cos∅ = p / Vi
cos∅ = 1500 / ( 380 × 5 ) = 0.7894
∅ = cos⁻¹ (0.7894)
∅ = 37.87°
Q = VIsin∅
Q = 380 × 5 × sin( 37.87° )
Q = 1.1663 KVAR
Therefore Reactive Power (kvar) is 1.1663 KVAR
c) Apparent Power (kVA)
S = P + jQ
= ( 1500 + J 1166.3 ) VA
S = 1900 ∠ 37.87° VA
S = 1.9 KVA
Therefore Apparent Power (kVA) is 1.9 KVA
d) Power Factor
p = V×i×cos∅
cos∅ = p / Vi
cos∅ = 1500 / ( 380 × 5 ) = 0.7894
Therefore the Power Factor cos∅ is 0.7894
e) The impedance Z in polar and rectangular form
Z = 380 / ( S∠-37.87) = V/I
Z = ( 60 + j 46.647) Ω
Z = 76 ∠ 37.87° Ω
Therefore the impedance Z in polar and rectangular form is 76 ∠ 37.87° Ω
Using the following data, determine the percentage retained, cumulative percentage retained, and percent passing for each sieve.
Sieve size Weight retained (g) No. 4 59.5 No. 8 86.5 No. 16 138.0 No. 30 127.8 No. 50 97.0 No. 100 66.8 Pan 6.3
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
Write out simple definitions in words and equations for the following:
a. a1
b. b1
c. S11
d. S12
e. S21
f. S22
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]
The seers were of the opinion that_____ . *
a healthy mind guides a healthy body.
the healthy body needs no exercise.
a healthy mind resides in a healthy body.
the healthy mind resides in every body.
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.
7. The surface finish for the cylinder walls usually depends on the
O A. type of engine oil used.
O B. sharpness of the cylinder bore edges.
O C.type of piston rings used
O D. cylinder wall-to-piston clearance.
A mixture of octane, C8H18, and air flowing into a combustor has 60% excess air and 1 kmol/s of octane. What is the mole flow rate (kmol/s) of CO2 in the product stream?
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
We put capacitors on our voltage supplies in order to filter out high frequency noise. Which is better. a 10uF capacitor or a 0.1uF capacitor? Why?
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
is a process that is used to systematically solve problems.
design
engineering
brainstorming
O teamwork
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.
In beams, why is the strain energy from bending moments much bigger than the strain energy from transverse shear forces? Choose one or more of the following options.
a) The stresses due to bending moments is much more than the stresses from transverse shear.
b) The strains due to bending moments is much more than the strains from transverse shear.
c) The deformations due to bending moments is much more than the deformations from transverse shear.
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.
What overall material composition would be required to give a material made up of 50wt% mullite and 50wt% alumina at 1400°C?
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}
A gas stream contains 18.0 mole% hexane and the remainder nitrogen. The stream flows to a condenser, where its temperature is reduced and some of the hexane is liquefied. The hexane mole fraction in the gas stream leaving the condenser is 0.0500. Liquid hexane condensate is recovered at a rate of 1.50 L/min.
(a) What is the flow rate of the gas stream leaving the condenser in mol/min? (Hint : First calculate the molar flow rate of the condensate and note that the rates at which C6H14 and N2 enter the unit must equal the total rates at which they leave in the two exit streams.)
(b) What percentage of the hexane entering the condenser is recovered as a liquid?
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 .
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:
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
If you make a mistake in polarity when measuring the value of DC voltage in a circuit with a digital VOM, what will happen? A. The meter will be damaged. B. The meter will read positive voltage only. C. The meter will display a negative sign. D. The meter will display OL which states an overload condition.
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
Air is compressed isothermally from 13 psia and 55°F to 80 psia in a reversible steady-flow device. Calculate the work required, in Btu/lbm, for this compression. The gas constant of air is R.
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
In a p+-n Si junction, the n side has a donor concentration of 1016 cm^-3. If ni = 1010 cm^-3, relative dielectric constant Pr = 12, calculate the depletion width at a reverse bias of 100 V? What is the electric field at the mid-point of the depletion region on the n side?
Answer:
This graph shows linear
y = f(x) and y = g(x).
Find the solution to the equation f(x) - g(x) = 0
A cylindrical specimen of Aluminium having a diameter of 12.8 mm and gauge length of 50.8 is pulled in tension. Use the data given below to:A) Plot the data as engineering stress versus engineering strain. B) Compute the modulus of elasticity. C) Determine the yield strength at a strain offset of 0.002. D) Determine the tensile strength of this alloy.E) What is the approximate ductility, in percent elongation?Load (N) Length0 50.8007330 50.85115100 50.90223100 50.95230400 51.00334400 51.05438400 51.30841300 51.81644800 52.83246200 53.84847300 54.86447500 55.88046100 56.89644800 57.65842600 58.42036400 59.182
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%
The structure of a house is such that it loses heat at a rate of 3800 kJ/h per C di erence between the indoors and outdoors. A heat pump that requires a power input of 4 kW is used to maintain this house at 24C. Determine the lowest outdoor temperature for which the heat pump can meet the heating requirements of this house.
Answer:
-9.5° C
Explanation:
See attachment for calculations.
On the concluding parts, from the attachment, we have that
√[(297000 * 4)/(1056)] = 297 - T(l), and solving further, we get
297 - T(l) = √(1188000/1056)
297 - T(l) = √1125
297 - T(l) = 33.5
T(l) = 297 - 33.5
T(l) = 263.5
When you convert back to °C, we have
263.5 - 273 = -9.5° C
A person holds her hand out of an open car window while the car drives through still air at 65 mph. Under standard atmospheric conditions, what is the maximum pressure on her hand? What would be the maximum pressure if the "car" were an Indy 500 racer traveling 200 mph?
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]
Un mol de gas ideal realiza un trabajo de 3000 J sobre su entorno, cuando se expande de manera isotermica a una temperatura de 58°C, cuando su volumen inicial es de 25 L. Determinar el volumen final
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
A rigid tank of volume of 0.06 m^3 initially contains a saturated mixture of liquid and vapor of H2O at a pressure of 15 bar and a quality of 0.2. The tank has a pressure-regulating venting valve that allows pressure to be constant. The tank is subsequently being heated until its content becomes a saturated vapor (of quality 1.0). During heating, the pressure-regulating valve keeps the pressure constant in the tank by allowing saturated vapor to escape. You can neglecting the kinetic and potential energy effects.
Required:
a. Determine the total mass in the tank at the initial and final states, in kg.
b. Calculate the amount of heat (in kJ) transferred from the initial state to the final state.
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
Which method of freezing preserves the quality and taste of food?
Answer:
commercial freezing
Explanation:
smaller ice crystals are formed this causes less damage to cell membranes so the quality is less effected
I need help with simply science
Answer:
mountain ranges may be
Which of the following is not one of the common classifications of product liability defects? A. Manufacture B. Materials C. Packaging D. Both "Materials" and "Packaging" E. Design
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
Oil with a kinematic viscosity of 4 10 6 m2 /s fl ows through a smooth pipe 12 cm in diameter at 2.3 m/s. What velocity should water?
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
If it is desired to lay off a distance of 10,000' with a total error of no more than ± 0.30 ft. If a 100' tape is used and the distance can be measured using full tape measures, what is the maxim error per tape measure allowed?
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
Which type of forming operation produces a higher quality surface finish, better mechanical properties, and closer dimensional control of the finished piece?A. Hot working.B. Cold working.
Answer:
Option B (Cold working) would be the correct alternative.
Explanation:
Cold working highlights the importance of reinforcing material without any need for heat through modifying its structure or appearance. Metal becomes considered to have been treated in cold whether it is treated economically underneath the material's transition temperature. The bulk of cold operating operations are carried out at room temperature.The other possibility isn't linked to the given scenario. Therefore the alternative above is the right one.
A roadway is to be designed on a level terrain. The roadway id 500 ft. Five cross-sections have been selected at 0 ft, 125 ft, 250 ft, 375 ft, and 500 ft. the cross sections have areas of 130 ft^2, 140 ft^2, 60 ft^2, 110 ft^2, and 120 ft^2. What is the volume needed along this road assuming a 6% shrinkage?
Answer:
51112.5 ft^3
Explanation:
Determine the volume needed along the road when we assume a 6% shrinkage
shrinkage factor = 1 - shrinkage = 1 - 0.06 = 0.94
first we have to calculate the volume between the cross sectional areas (i.e. A1 ---- A5 ) using average end area method
Volume between A1 - A2
= (125 ft - 0 ft) * [(130 ft^2 + 140 ft^2) / 2]
= 125 ft * 135 ft^2
= 16875 ft^3
Volume between A2 - A3
= (250 ft - 125 ft) * [(140 ft^2 + 60 ft^2) / 2]
= 125 ft * (200 ft^2 / 2)
= 12500 ft^3
Volume between A3 - A4
= (375 ft - 250 ft) * [(60 ft^2 + 110 ft^2) / 2]
= 125 ft * (170 ft^2 / 2)
= 10625 ft^3
Volume between A4 - A5
(500 ft - 375 ft) * [(110 ft^2 + 120 ft^2) / 2]
= 125 ft * 115 ft^2
= 14375 ft3
Hence the total volume along the 500 ft road
= ∑ volumes between cross sectional areas
= 16875 ft^3 + 12500 ft^3 + 10625 ft^3 + 14375 ft^3 = 54375 ft^3
Finally the volume needed along this road is calculated as
Total volume * shrinkage factor
= 54375 * 0.94 = 51112.5 ft^3
Given a 12-bit A/D converter operating over a voltage range from ????5 V to 5 V, how much does the input voltage have to change, in general, in order to be detectable
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
Consider a pan of water being heated (a) by placing it on an electric range and (b) by placing a heating element in the water. Which method is a more efficient way of heating water? Explain.
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.
Water leaves a penstock (the flow path through a hydroelectric dam) at a velocity of 100 ft/s. How deep is the water behind the dam (in ft). Neglect friction. [h = 155 ft]
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
What is difference between a backdoor, a bot, a keylogger, and psyware,a nd a rootkit? Can they all present in the same malware?
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.