Answer:
Explanation:
B. you would grab the plug closest to the outlet
If you don't have enough experience, it's always best to leave socket changing to the experts. If you make a mistake, you might inflict harm and potentially endanger yourself and other people. Read on if you're interested in learning how to change a socket safely. Thus, option D is correct.
What, removing an electrical plug from its socket?Grip the plug, not the electrical cable, when taking an electrical plug out of its socket. Before handling an electrical switch, socket, or outlet, hands must be fully dry.
Reduce the extra so that it rests only on top of the existing plasterboard. If necessary, push it back a little by using your finger. Fill the dent with ready-mixed filler or powdered filler, whichever you want, and bring it flush with the surrounding wall. Allow to dry, then sand off any excess.
Therefore, It is acceptable to grasp the electrical cord when removing an electrical plug from its socket
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g The inside surface of a 17 mm inner diameter tube with a 2.4 mm thick wall indicates a temperature of 46 deg C. The outside temperature is 43 deg C. The tube is 5 m long. If the tube material has a conductivity of 0.15 W/m/K, estimate the heat transfer rate through the tube wall assuming SS 1D conduction. Indicate the direction of heat transfer with a or - sign ( meaning outward and vice versa). Express your answer using two significant digits in W.
Answer:
-50 W
Explanation:
The heat transfer rate Q = kA(T₂ - T₁)/d where k = thermal conductivity of material = 0.15 W/m-K, A = surface area of tube = πdL where d = diameter of tube = 17 mm = 0.017 m and L = length of tube = 5 m, T₁ = inside temperature = 46 °C, T₂ = outside temperature = 43 °C and d = thickness of tube = 2.4 mm = 0.0024 m
Since Q = kA(T₂ - T₁)/d ,
Q = kπdL(T₂ - T₁)/d
substituting the values of the variables into the equation, we have
Q = 0.15 W/m-K × π × 0.017 m × 5 m(43 °C - 46 °C )/0.0024 m
Q = 0.01275π Wm/K(-3 K )/0.0024 m
Q = -0.03825π Wm/0.0024 m
Q = -0.1202 Wm/0.0024 m
Q = -50.07 W
Q = -50 W
So, the heat transfer rate is -50 W meaning heat transfer out of the tube.
If there is a discrepancy between Chick-fil-A food safety requirements and local Health Department
regulations, what should Team Member do?
The following should be done by the team member:
It is important to follow both Chick-fil-A food safety requirements and local Health Department regulations. In the case when there is a discrepancy between the two, always follow the more stringent requirement. Any other appearance or grooming issue not covered in these materials may be addressed at the discretion of the Operator.
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A levee will be constructed to provide some flood protection for a residential area. The residences are willing to accept a one-in-five chance that the levee will be overtopped in the next 15 years. Assuming that the annual peak streamflow follows a lognormal distribution with a log10(Q[ft3/s]) mean and standard deviation of 1.835 and 0.65 respectively, what is the design flow in ft3/s?
Answer:
1709.07 ft^3/s
Explanation:
Annual peak streamflow = Log10(Q [ft^3/s] )
mean = 1.835
standard deviation = 0.65
Probability of levee been overtopped in the next 15 years = 1/5
Determine the design flow ins ft^3/s
P₁₅ = 1 - ( q )^15 = 1 - ( 1 - 1/T )^15 = 0.2
∴ T = 67.72 years
Q₁₅ = 1 - 0.2 = 0.8
Applying Lognormal distribution : Zt = mean + ( K₂ * std ) --- ( 1 )
K₂ = 2.054 + ( 67.72 - 50 ) / ( 100 - 50 ) * ( 2.326 - 2.054 )
= 2.1504
back to equation 1
Zt = 1.835 + ( 2.1504 * 0.65 ) = 3.23276
hence:
Log₁₀ ( Qt(ft^3/s) ) = Zt = 3.23276
hence ; Qt = 10^3.23276
= 1709.07 ft^3/s
If a heat engine has an efficiency of 30% and its power output is 600 W, what is the rate of heat input from the combustion phase
Answer:
The heat input from the combustion phase is 2000 watts.
Explanation:
The energy efficiency of the heat engine ([tex]\eta[/tex]), no unit, is defined by this formula:
[tex]\eta = \frac{\dot W}{\dot Q}[/tex] (1)
Where:
[tex]\dot Q[/tex] - Heat input, in watts.
[tex]\dot W[/tex] - Power output, in watts.
If we know that [tex]\dot W = 600\,W[/tex] and [tex]\eta = 0.3[/tex], then the heat input from the combustion phase is:
[tex]\eta = \frac{\dot W}{\dot Q}[/tex]
[tex]\dot Q = \frac{\dot W}{\eta}[/tex]
[tex]\dot Q = \frac{600\,W}{0.3}[/tex]
[tex]\dot Q = 2000\,W[/tex]
The heat input from the combustion phase is 2000 watts.
The following is a correlation for the average Nusselt number for natural convection over spherical surface. As can be seen in the above, the Nusselt number approaches 2 as Rayleigh number approaches zero. Prove that this situation corresponds to conduction heat transfer and in conduction heat transfer over sphere, the Nusselt number becomes 2. Hint: First step: Write an expression for heat transfer between two spherical shells that share the same center. Second step: Assume the outer spherical shell is infinitely large.
Answer:
Explanation:
[tex]r_2=[/tex]∞
[tex]q=4\pi kT_1(T_2-T_1)\\[/tex]
[tex]q=2\pi kD.[/tex]ΔT--------(1)
[tex]q=hA[/tex] ΔT[tex]=4\pi r_1^2(T_2_s-T_1_s)\\[/tex]
[tex]N_u=\frac{hD}{k} = 2+\frac{0.589 R_a^\frac{1}{4} }{[1+(\frac{0.046}{p_r}\frac{9}{16} )^\frac{4}{9} } ------(3)[/tex]
By equation (1) and (2)
[tex]2\pi kD.[/tex]ΔT=h.4[tex]\pi r_1^2[/tex]ΔT
[tex]2kD=hD^2\\\frac{hD}{k} =2\\N_u=\frac{hD}{k}=2\\[/tex]-------(4)
From equation (3) and (4)
So for sphere [tex]R_a[/tex]→0
A multipurpose transformer has a secondary coil with several points at which a voltage can be extracted, giving outputs of 5.60, 12.0, and 480 V. (a) The input voltage is 220 V to a primary coil of 230 turns. What are the numbers of turns in the parts of the secondary used to produce the output voltages
Answer:
Explanation:
A multipurpose transformer can act as step up as well as step down transformer according to the desired setting by a user.
When the voltage at the output is greater than the voltage at the input of the transformer then it acts as step-up transformer and vice-versa acting is a step down transformer.
Given that:
input (primary) voltage of the transformer, [tex]V_i=220~V[/tex]
no. of turns in the primary coil, [tex]N_i=230[/tex]
When the output voltage is 5.60 V:[tex]V_o=5.60~V[/tex]
[tex]\frac{N_i}{N_o} =\frac{V_i}{V_o}[/tex]
[tex]\frac{N_o}{230}=\frac{5.60}{220}[/tex]
[tex]N_o=5.85\approx 6[/tex] turns compensating the losses
When the output voltage is 12.0 V:[tex]V_o=12.0~V[/tex]
[tex]\frac{N_i}{N_o} =\frac{V_i}{V_o}[/tex]
[tex]\frac{N_o}{230}=\frac{12.0}{220}[/tex]
[tex]N_o=12.45\approx 13[/tex] turns compensating the losses
When the output voltage is 480 V:[tex]V_o=480~V[/tex]
[tex]\frac{N_i}{N_o} =\frac{V_i}{V_o}[/tex]
[tex]\frac{N_o}{230}=\frac{480}{220}[/tex]
[tex]N_o=501.8\approx 502[/tex] turns compensating the losses
(50 POINTS) How many people use pipes in the world? How do you know this?
Answer:
7.9 billion people
Explanation:
Page Title
Submit
Add a pair of radio buttons to your form, each nested in its own label element.
One should have the option of car and the other should have the option of bike.
Both should share the name attribute of “vehicle” to create a radio group
Make sure the radio buttons are nested with the form
Make sure that the name attributes appear after the type
Answer:
The code is as follows:
<form name = "myForm">
<div>
<input type="radio" name="vehicle" value="D0" id="D0"/>
<label for="D0">Car</label>
</div>
<div>
<input type="radio" name="vehicle" value="D1" id="D1"/>
<label for="D1">Bike</label>
</div>
</form>
Explanation:
This defines the first button
<input type="radio" name="vehicle" value="D0" id="D0"/>
<label for="D0">Car</label>
This defines the second button
<input type="radio" name="vehicle" value="D1" id="D1"/>
<label for="D1">Bike</label>
The code is self-explanatory, as it follows all the required details in the question
Technician A says that a graphing multi-meter may be used to verify signals going to and from electrical and electronic components. Technician B says that digital storage oscilloscope may be used to verify signals going to and from electrical and electronic components. Who is correct
Answer:
Both are correct.
Explanation:
Graphing multi meter is used to verify signals that move from electrical components. Digital oscilloscope is an equipment which stores and analyzes input signals with digital technique.
it is a small sharp and printed item for fine worker in trimming scallops clipping threads and cutting large eyelets
Answer:
embroidery scissor
Explanation:
is small, sharp and pointed good for fine work use trimming scallops,clipping threads,and cutting large eyelets.
hope this helps
6. When the engine stalls or the power unit fails, on a car with power
brakes, the service brake pedal will
A. Take about the same amount of pressure
B. Take more pressure to stop
C. Take less pressure to stop
D. Become locked in place and no longer help stop the car
1. Add:
(i) 5xy, -2xy, -11xy, 8xy
(iv) 3a - 2b + c, 5a + 8b -70
Answer:
(i) 0
(iv) 8a+6b+c-70
Explanation:
Hope this helps you
A start-up is expanding overseas and spends an excessive amount of time on recruiting and hiring activities, hindering its ability to focus on the core aspects of its business. How can a Human Capital Management (HCM) platform provider benefit this company?
Answer:
Human Capital Management (HCM) will help the start-up firm manage its recruiting and hiring activities.
Explanation:
Human Capital Management (HCM) Platform will assist the start-up firm manage its main point of access by keeping the employee records and maintaining the wages and salaries, managing the benefits, time, and attendance, and carrying out performance reviews including looking after the most important asset employees.
For a steel alloy it has been determined that a carburizing heat treatment of 3-h duration will raise the carbon concentration to 0.38 wt% at a point 2.6 mm from the surface. Estimate the time (in h) necessary to achieve the same concentration at a 6.1 mm position for an identical steel and at the same carburizing temperature.
Answer:
The right answer is "16.5 hrs".
Explanation:
Given values are:
[tex]x_1=2.6 \ mm[/tex]
[tex]t_1=3 \ hrs[/tex]
[tex]x_2=6.1 \ mm[/tex]
As we know,
⇒ [tex]\frac{x^2}{Dt}=constant[/tex]
or,
⇒ [tex]\frac{x_1^2}{t_1} =\frac{x_2^2}{t_2}[/tex]
⇒ [tex]t_2=(\frac{x_2}{x_1})^2\times t_1[/tex]
By putting the values, we get
[tex]=(\frac{6.1}{2.6} )^2\times 3[/tex]
[tex]=5.5\times 3[/tex]
[tex]=16.5 \ hrs[/tex]
Suppose a causal CT LTI system has bilateral Laplace transform H(s) 2s - 2 $2 + (10/3)s + 1 (8)
(a) Write the linear constant coefficient differential equation (LCCDE) relating a general input x(t) to its corresponding output y(t) of the system corresponding to this transfer function in equation (8).
(b) Suppose the input x(t) = e-tu(t). Find the output y(t). In part (c), the output signal can be expressed as y(t) = - e-(1/3)t u(t) + e-tu(t) e-3tu(t), - 019 Where a, b, and care positive integers. What are they? a = b = C=
Solution :
Given :
[tex]$H(S) =\frac{2S-2}{S^2+\left(\frac{10}{3}\right) S+1}$[/tex]
Transfer function, [tex]$H(S) =\frac{Y(S)}{K(S)}= \frac{2S-2}{S^2+\left(\frac{10}{3}\right) S+1}$[/tex]
[tex]$Y(S) \left(S^2+\frac{10}{3}S+1\right) = (2S-2) \times (S)$[/tex]
[tex]$S^2Y(S) + \frac{10}{3}(SY(S)) + Y(S) = 2(S \times (S)) - 2 \times (S)$[/tex]
Apply Inverse Laplace Transforms,
[tex]$\frac{d^2y(t)}{dt^2} + \frac{10}{3} \frac{dy(t)}{dt} + y(t)=2 \frac{dx(t)}{dt} - 2x(t)$[/tex]
The above equation represents the differential equation of transfer function.
Given : [tex]$x(t)=e^{-t} u(t) \Rightarrow X(S) = \frac{1}{S+1}$[/tex]
We have : [tex]$H(S) =\frac{Y(S)}{K(S)}= \frac{2S-2}{S^2+\left(\frac{10}{3}\right) S+1}$[/tex]
[tex]$Y(S) = X(S) \times \frac{6S-6}{3S^2+10 S + 3} = \frac{6S-6}{(S+1)(3S+1)(S+3)}$[/tex]
[tex]$Y(S) = \frac{A}{S+1}+\frac{B}{3S+1} + \frac{C}{S+3}[/tex]
[tex]$A = Lt_{S \to -1} (S+1)Y(S)=\frac{6S-6}{(3S+1)(S+3)} = \frac{-6-6}{(-3+1)(-1+3)} = 3$[/tex]
[tex]$B = Lt_{S \to -1/3} (3S+1)Y(S)=\frac{6S-6}{(S+1)(S+3)} = \frac{-6/3-6}{(1/3+1)(-1/3+3)} = \frac{-9}{2}$[/tex]
[tex]$C = Lt_{S \to -3} (S+3)Y(S)=\frac{6S-6}{(S+1)(3S+1)} = \frac{-18-6}{(-3+1)(-9+1)} = \frac{-3}{2}$[/tex]
So,
[tex]$Y(S) = \frac{3}{S+1} - \frac{9/2}{3S+1} - \frac{3/2}{S+3}$[/tex]
[tex]$=\frac{3}{S+1} - \frac{3/2}{S+1/3} - \frac{3/2}{S+3}$[/tex]
Applying Inverse Laplace Transform,
[tex]$y(t) = 3e^{-t}u(t)-\frac{3}{2}e^{-t/3}u(t) - \frac{3}{2}e^{-3t} u(t)$[/tex]
[tex]$=\frac{-3}{2}e^{-\frac{1}{3}t}u(t) + \frac{3}{1}e^{-t}u(t)-\frac{3}{2}e^{-3t} u(t)$[/tex]
where, a = 2
b = 1
c= 2
3-71A 20mm diameter steel bar is to be used as a torsion spring. If the torsional stress in the bar is not to exceed 110 MPa when one end is twisted through an angle of 15 degrees, what must be the length of the bar
Answer:
The right answer is "1.903 m".
Explanation:
Given that,
[tex]\tau =110 \ MPa[/tex]
[tex]G=80 \ GPa[/tex]
[tex]\Theta=15\times \frac{\pi}{180}[/tex]
[tex]=\frac{\pi}{12}[/tex]
[tex]d=20 \ mm[/tex]
As we know,
⇒ [tex]\frac{\tau}{r}=\frac{G \Theta}{L}[/tex]
Or,
⇒ [tex]L=\frac{G \theta r}{\tau}[/tex]
[tex]=\frac{80\times 10^3}{110}\times \frac{\pi}{12}\times 10[/tex]
[tex]=1903.9 \ mm[/tex]
or,
[tex]=1.903 \ m[/tex]
Steam enters an adiabatic turbine at 6 MPa, 600°C, and 80 m/s and leaves at 50 kPa, 100°C, and 140 m/s. If the power output of the turbine is 5 MW, determine (a) the reversible power output and (b) the second-law efficiency of the turbine. Assume the surroundings to be at 25°C.
Answer:
(a) the reversible power output of turbine is 5810 kw
(b) The second-law efficiency of he turbine = 86.05%
Explanation:
In state 1: the steam has a pressure of 6 MPa and 600°C. Obtain the enthalpy and entropy at this state.
h1 = 3658 kJ/kg s1=7.167 kJ/kgK
In state 2: the steam has a pressure of 50 kPa and 100°C. Obtain the enthalpy and entropy at this state
h2 = 2682kl/kg S2= 7.694 kJ/kg
Assuming that the energy balance equation given
Wout=m [h1-h2+(v1²-v2²) /2]
Let
W =5 MW
V1= 80 m/s V2= 140 m/s
h1 = 3658kJ/kg h2 = 2682 kJ/kg
∴5 MW x1000 kW/ 1 MW =m [(3658-2682)+ ((80m/s)²-(140m/s)²)/2](1N /1kg m/ s²) *(1KJ/1000 Nm)
m = 5.158kg/s
Consider the energy balance equation given
Wrev,out =Wout-mT0(s1-s2)
Substitute Wout =5 MW m = 5.158kg/s 7
s1= 7.167 kJ/kg-K s2= 7.694kJ/kg-K and 25°C .
Wrev,out=(5 MW x 1000 kW /1 MW) -5.158x(273+25) Kx(7.167-7.694)
= 5810 kW
(a) Therefore, the reversible power output of turbine is 5810 kw.
The given values of quantities were substituted and the reversible power output are calculated.
(b) Calculating the second law efficiency of the turbine:
η=Wout/W rev,out
Let Wout = 5 MW and Wrev,out = 5810 kW
η=(5 MW x 1000 kW)/(1 MW *5810)
η= 86.05%
Use pseudocode. 1) Prompt for and input a saleswoman's sales for the month (in dollars) and her commission rate (percentage). Output her commission for that month. Note that you will need the following Variables: SalesAmount CommissionRate CommissionEarned
You will need the following formula: CommissionEarned= Sales Amount * (commissionrate/100)
Answer:
The pseudocode is as follows:
Input SalesAmount
Input CommissionRate
CommissionEarned= SalesAmount * (CommissionRate/100)
Print CommissionEarned
Explanation:
This gets input for SalesAmount
Input SalesAmount
This gets input for CommissionRate
Input CommissionRate
This calculates the CommissionEarned
CommissionEarned= SalesAmount * (CommissionRate/100)
This prints the calculated CommissionEarned
Print CommissionEarned
A circular rod with a gage length of 3.1 m and a diameter of 3 cm is subjected to an axial load of 68 kN . If the modulus of elasticity is 200 GPa , what is the change in length
Answer:
1.49 mm
Explanation:
The modulus of elasticity, Y = stress/strain = σ/ε
σ = F/A where F = load = 68 kN = 68 × 10³ N and A = cross-sectional area of rod = πd²/4 where d = diameter of rod = 3 cm = 3 × 10⁻² m.
ε = ΔL/L where ΔL = change in length of the circular rod and L = length of circular rod = 3.1 ,
So, Y = σ/ε
Y = F/A ÷ ΔL/L
Y = FL/AΔL
making the change in length ΔL subject of the formula, we have
ΔL = FL/AY
substituting the value of A into the equation, we have
So, ΔL = FL/(πd²/4)Y
ΔL = 4FL/πd²Y
Since Y = 200 GPa = 200 × 10⁹ Pa
Substituting the values of the variables into the equation, we have
ΔL = 4FL/πd²Y
ΔL = 4 × 68 × 10³ N × ×3.1 m/[π(3 × 10⁻²m)² × 200 × 10⁹ Pa]
ΔL = 843.2 × 10³ Nm/[9π × 10⁻⁴m² × 200 × 10⁹ Pa]
ΔL = 843.2 × 10³ Nm/[1800π × 10⁵ N]
ΔL = 843.2 × 10³ Nm/5654.87 × 10⁵ N
ΔL = 0.149 × 10⁻² m
ΔL = 1.49 × 10⁻³ m
ΔL = 1.49 mm
The change in length of the circular rod is 1.49 mm
Air is compressed in a well insulated compressor from 95 kPa and 27 C to 600 kPa and 277 C. Use the air tables; assume negligible changes in kinetic and potential energy. Find the isentropic efficiency of the compressor. Find the exit temperature of the air if the compressor was reversible.
Answer:
a) 1.9%
b) T2s = 505.5 k = 232.5°C
Explanation:
P1 = 95 kPa
T1 = 27°C = 300 k
P2 = 600 kPa
T1 = 277°c = 550 k
Table used : Table ( A - 17 ) Ideal gas properties of air
a) determining the isentropic efficiency of the compressor
Л = ( h2s - h1 ) / ( h2a - h1 ) ---- ( 1 )
where ; h1 = 300.19 kJ/kg , T1 = 300 K , h2a = 554.74 kJ/kg , T2 = 550 k
To get h2s we have to calculate the the value of Pr2 using Pr1(relative pressure)
Pr2 = P2/P1 * Pr = ( 600 / 95 ) * 1.306 hence; h2s = 500.72 kJ/kg
back to equation1
Л = 0.019 = 1.9%
b) Calculate the exit temperature of the air if compressor is reversible
if compressor is reversible the corresponding exit temperature
T2s = 505.5 k = 232.5°C
given that h2s = 500.72 kJ/kg
Imagine a cantilever beam fixed at one end with a mass = m and a length = L. If this beam is subject to an inertial force and a uniformly distributed load = w, what is the moment present at a length of L/4?
Answer:
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Explanation:
j
grudba) Complete the following methods description using the correct tense for the verb in brackets. (This student is using passive voice rather than any human agents at the request of the instructor.) Student Lab Report Identical tensile test procedures were performed on all test specimens. Each of the metal specimens ____1____ [have] an indentation near the center to ensure that the fracture point would occur in this region. Tension tests ____2____ [conduct] as follows. Two pieces of reflective tape ____3____ [place] approximately 1 inch apart in the center of the specimen where the indentation 4 [locate]. The width and the thickness of the specimen at this location _____5_____ [measure] using a Vernier caliper. Then the specimen _____6____ [secure] in the MTS Load Frame. A laser extensometer _____7_____ [place] into position to measure the deformation of the specimen. The laser extensometer ______8_ __ [use] to measure the original distance between the pieces of reflective tape. The MTS ________9____ [set] to elongate the specimen one tenth of an inch every minute.
Answer:
Each of the metal specimens HAS an indentation near the center to ensure that the fracture point would occur in this region. Tension tests WERE CONDUCTED as follows. Two pieces of reflective tape WERE PLACED approximately 1 inch apart in the center of the specimen where the indentation 4 WAS LOCATED. The width and the thickness of the specimen at this location WAS MEASURED using a Vernier caliper. Then the specimen WAS SECURED in the MTS Load Frame. A laser extensometer WAS PLACED into position to measure the deformation of the specimen. The laser extensometer WAS USED to measure the original distance between the pieces of reflective tape. The MTS WAS SET to elongate the specimen one tenth of an inch every minute.
ow Pass Filter Design 0.0/5.0 points (graded) Determine the transfer function H(s) for a low pass filter with the following characteristics: a cutoff frequency of 100 kHz a stopband attenuation rate of 40 dB/decade. a nominal passband gain of 20 dB, which drops to 14 dB at the cutoff frequency Write the formula for H(s) that satisfies these requirements:
Answer:
H(s) = 20 / [ 1 + s / 10^5 ]^2
Explanation:
Given data:
cutoff frequency = 100 kHz
stopband attenuation rate = 40 dB/decade
nominal passband gain = 20 dB
new nominal passband gain at cutoff = 14 dB
Represent the transfer function H(s)
The attenuation rate show that there are two(2) poles
H(s) = k / [ 1 + s/Wc ]^2 ----- ( 1 )
where : Wc = 100 kHz = 10^5 Hz , K = 20 log k = 20 dB ∴ k = 20
Input values into equation 1
H(s) = 20 / [ 1 + s / 10^5 ]^2
Water steam enters a turbine at a temperature of 400 o C and a pressure of 3 MPa. Water saturated vapor exhausts from the turbine at a pressure of 125 kPa. At steady state, the work output of the turbine is 530 kJ/kg. The surrounding air is at 20 o C. Neglect the changes in kinetic energy and potential energy. Determine (20 points) (a) the heat transfer from the turbine to the surroundings per unit mass flow rate, (b) the entropy generation during this process.
Answer:
a) -505.229 kJ/Kg
b) -1.724 kJ/kg
Explanation:
T1 = 400°C
P1 = 3 MPa
P2 = 125 kPa
work output = 530 kJ/kg
surrounding temperature = 20°C = 293 k
A) Calculate heat transfer from Turbine to surroundings
Q = h2 + w - h1
h ( enthalpy )
h1 = 3231.229 kj/kg
enthalpy at P2
h2 = hg = 2676 kj/kg
back to equation 1
Q = 2676 + 50 - 3231.229 = -505.229 kJ/Kg ( i.e. heat is lost )
b) Entropy generation
entropy generation = Δs ( surrounding ) + Δs(system)
= - 505.229 / 293 + 0
= -1.724 kJ/kg
Alice and Bob both have RSA Public-Private key pairs: (PUA, PRA) and (PUB, PRB). They also have cryptographic functions E_AES / D_AES to encrypt / decrypt using AES; and E_RSA and D_RSA to encrypt / decrypt using RSA. Alice wants to sent a high resolution video of a large secret facility to Bob.
A. Show how Alice can securely and efficiently send the video to Bob. You are required to use the cryptographic functions above to get full credit;
B. Does your solutions assure confidentiality? How / Why not?
C. Does your solutions assure non-repudiation? How / Why not?
D. Does your solutions assure integrity? How / Why not?
E. Does your solutions assure replay attacks? How / Why not?
Solution :
B. yes, the given solution assures confidentiality. The sender Alice encrypting his messages with its own private key PRA which provides authentication. Sender Alice further encrypts his messages with the receiver's public key PUB provides confidentiality.
C. So the given solution provides non repudiation. Alice and Bob who are exchanging messages. In one case, Alice denies sending a messages to Bob that he claims to have received being able to counter Alice's denial is caused non repudiation of origin.
D. The given solution provides integrity. Because it provides authentication and have not been changed.
E. It does not provide replay attacks because it does not captures the traffic. The client does not receive the messages twice.
Steam at 4 MPa and 350°C is expanded in an adiabatic turbine to 125kPa. What is the isentropic efficiency (percent) of this turbine if the steam is exhausted as a saturated vapor?
Answer:
[tex]\eta_{turbine} = 0.603 = 60.3\%[/tex]
Explanation:
First, we will find actual properties at given inlet and outlet states by the use of steam tables:
AT INLET:
At 4MPa and 350°C, from the superheated table:
h₁ = 3093.3 KJ/kg
s₁ = 6.5843 KJ/kg.K
AT OUTLET:
At P₂ = 125 KPa and steam is saturated in vapor state:
h₂ = [tex]h_{g\ at\ 125KPa}[/tex] = 2684.9 KJ/kg
Now, for the isentropic enthalpy, we have:
P₂ = 125 KPa and s₂ = s₁ = 6.5843 KJ/kg.K
Since s₂ is less than [tex]s_g[/tex] and greater than [tex]s_f[/tex] at 125 KPa. Therefore, the steam is in a saturated mixture state. So:
[tex]x = \frac{s_2-s_f}{s_{fg}} \\\\x = \frac{6.5843\ KJ/kg.K - 1.3741\ KJ/kg.K}{5.91\ KJ/kg.K}\\\\x = 0.88[/tex]
Now, we will find [tex]h_{2s}[/tex](enthalpy at the outlet for the isentropic process):
[tex]h_{2s} = h_{f\ at\ 125KPa}+xh_{fg\ at\ 125KPa}\\\\h_{2s} = 444.36\ KJ/kg + (0.88)(2240.6\ KJ/kg)\\h_{2s} = 2416.088\ KJ/kg[/tex]
Now, the isentropic efficiency of the turbine can be given as follows:
[tex]\eta_{turbine} = \frac{h_1-h_2}{h_1-h_{2s}}\\\\\eta_{turbine} = \frac{3093.3\ KJ/kg-2684.9\ KJ/kg}{3093.3\ KJ/kg-2416.088\ KJ/kg}\\\\\eta_{turbine} = \frac{408.4\ KJ/kg}{677.212\ KJ/kg}\\\\\eta_{turbine} = 0.603 = 60.3\%[/tex]
Consider the equation y = 10^(4x). Which of the following statements is true?
A plot of log(y) vs. x would be linear with a slope of 4.
A plot of log(y) vs. log (x) would be linear with a slope of 10.
A plot of log(y) vs. x would be linear with a slope of 10.
A plot of y vs. log(x) would be linear with a slope of 4.
A plot of log(y) vs. log (x) would be linear with a slope of 4.
A plot of y vs. log(x) would be linear with a slope of 10.
Answer: Plot of [tex]\log y[/tex] vs [tex]x[/tex] would be linear with a slope of 4.
Explanation:
Given
Equation is [tex]y=10^{4x}[/tex]
Taking log both sides
[tex]\Rightarrow \log y=4x\log (10)\\\Rightarrow \log y=4x[/tex]
It resembles with linear equation [tex]y=mx+c[/tex]
Here, slope of [tex]\log y[/tex] vs [tex]x[/tex] is 4.
An industrial load with an operating voltage of 480/0° V is connected to the power system. The load absorbs 120 kW with a lagging power factor of 0.77. Determine the size of the capacitor in vars that is necessary to raise the power factor to 0.9 lagging.
Answer:
[tex]Q=41.33 KVAR\ \\at\\\ 480 Vrms[/tex]
Explanation:
From the question we are told that:
Voltage [tex]V=480/0 \textdegree V[/tex]
Power [tex]P=120kW[/tex]
Initial Power factor [tex]p.f_1=0.77 lagging[/tex]
Final Power factor [tex]p.f_2=0.9 lagging[/tex]
Generally the equation for Reactive Power is mathematically given by
Q=P(tan \theta_2-tan \theta_1)
Since
[tex]p.f_1=0.77[/tex]
[tex]cos \theta_1 =0.77[/tex]
[tex]\theta_1=cos^{-1}0.77[/tex]
[tex]\theta_1=39.65 \textdegree[/tex]
And
[tex]p.f_2=0.9[/tex]
[tex]cos \theta_2 =0.9[/tex]
[tex]\theta_2=cos^{-1}0.9[/tex]
[tex]\theta_2=25.84 \textdegree[/tex]
Therefore
[tex]Q=P(tan 25.84 \textdegree-tan 39.65 \textdegree)[/tex]
[tex]Q=120*10^3(tan 25.84 \textdegree-tan 39.65 \textdegree)[/tex]
[tex]Q=-41.33VAR[/tex]
Therefore
The size of the capacitor in vars that is necessary to raise the power factor to 0.9 lagging is
[tex]Q=41.33 KVAR\ \\at\\\ 480 Vrms[/tex]
Lab 5A Problem Input two DWORD values from the keyboard. Determine which number is larger or if they are even. Your program should look like the following: First number larger Enter a number 12 Enter a number 10 12 is the larger number Press any key to close this window... Second number larger Enter a number 10 Enter a number 12 12 is the larger number Press any key to close this window... Numbers Equal Enter a number 12 Enter a number 12 Numbers are equal Press any key to close this window...
Answer:
Explanation:
#include<iostream>
using namespace std;
int main()
{
int n1,n2;
cout<<"Enter a number:"<<endl; //Entering first number
cin>>n1;
cout<<"Enter a number:"<<endl; //Entering second number
cin>>n2;
if(n1%2==0 && n1%2==0) //Checking whether the two number are even or not
{
if(n1>n2)
{
cout<<n1<<" is the larger number"<<endl;
}
else if(n1==n2)
{
cout<<"Numbers are equal"<<endl;
}
else
{
cout<<n2<<" is the larger number"<<endl;
}
}
else
{
cout<<"The number are not even"<<endl;
}
}
Cite another example of information technology companies pushing the boundaries of privacy issues; apologizing, and then pushing again once scandal dies down. As long as the controversy fades, is there anything unethical about such a strategy?
Answer:
Explanation:
Tech Social Media giant FB is one of those companies. Not long ago the ceo was brought to court to accusations that his company was selling user data. Turns out this is true and they are selling their users private data to companies all over the word. Once the news turned to something else, people focused on something new but the company still continues to sell it's users data the same as before. This is completely unethical as the information belongs to the user and they are not getting anything while the corporation is profiting.
