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?

Answer:

True

Explanation:

Yes it is true that the Temperature gradient would also decrease with magnitude just as the distances rise from the centre line.

We have this cylinder equation as

[T1-T2 / ln(r1-r2)]2πKL

The radial distance is r2-r1

The gradient of temperature is T1-T2

From the equation,

The temperature gradient has a direct and proportional relationship to radial distance

T1-T2 ∝ ln(r2-r1)

1/T1-T2 = k(r2-r1)

This inverse relationship above confirms that the statement is true

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]

Answer:

can you please give us the option for this question

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]

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

Answer:

1. Drawing grid.

2. Orthogonal.

3. Geometries.

4. CTRL+N.

5. Thirteen (13).

Explanation:

CAD is an acronym for computer aided design and it is typically used for designing the graphical representation of a building plan. An example of a computer aided design (CAD) software is AutoCAD.

Some of the features of an AutoCAD software are;

1. Drawing grid: is a regular pattern of dots displayed on the screen of an AutoCAD software, which acts as a visual aid and it's also used to define the extent of a drawing.

2. Ortho is short or an abbreviation for orthogonal, which means either vertical or horizontal.

3. Tangent is a point where two geometries meet at just a single point.

4. If you want to create a new drawing, simply press CTRL+N for the short cut key.

5. There are thirteen object snaps (Osnap) that can help you perform your task on AutoCAD easily. The 13 object snaps (Osnap) are; Endpoint, Midpoint, Apparent intersect, Intersection, Quadrant, Extension, Tangent, Center, Insert, Perpendicular, Node, Parallel, and Nearest.

Answer:

H(s) =  10 / [ 1 + s / (200*10^3π ) ]^2

Explanation:

Characteristics of the high pass filter

Cutoff frequency = 100 kHz

stopband attenuation rate = 40 dB/decade

nominal passband gain = 20dB  = 20logK = 20

Formula for H(s) satisfying the requirements above

given that the stopband attenuation = 40 dB/decade the formula for H(s) that will satisfy the requirements is a second order filter

H(s) = K / ( 1 + s/Wo ) ^2  ----- ( 1 )

Wo = 2πf = 2π ( 100 * 10^3 ) =  200 * 10^3 π

 K = 10

back to equation ( 1 )

H(s) =  10 / [ 1 + s / (200*10^3π ) ]^2

Answer:

False

Explanation:

Bainite is a type of steel. It is formed by the decomposition of austenite. This happens at a temperature above MS but given that MS temperature is below the one where fine pearlite is formed.

In other words, when iron goes through a metastable crystallization phase Bainite is formed. It takes the rapid cooling, or quenching, of austenite for that to happen.

Metastability is the phenomenon where matter remains in an apparent state of equilibrium or stability even though it can achieve a more stable state. It can also be referred to as the condition of remaining in that pseudo unstable state for a very protracted period of time.

Bainite steel is used for the construction of components that require frequent usage and where plasticity, or tendency for deformity, as well as wear, cannot be tolerated.

Cheers

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.

Answer:

≈ 0.516 m^3/hr

Explanation:

Inlet of compressor = 200 SCMH

sheer standard conditions = 1 atm and 288.5 K

For oxygen :

critical pressure(Pc) = 49.8 atm

critical temperature Tc = 154.6 K

hence at compressor inlet

Tr = T / Tc = 288.5/154.6 = 1.866

Pr = P / Pc = 1 / 49.8 = 0.0204

Z1 ( from compressibility chart ) = 0.98

at compressor outlet

P2 = 500 bar = 500*0.9869 = 493.45 atm  , T2 = 360 k

hence : Pr = P / Pc = 493.45 / 49.8 = 9.91

            Tr = T / Tc = 360 / 154.6 = 2.33

Z2 ( from compressibility chart ) ≈ 1

V2( volumetric flow rate ) = V1*(P₁Z₂T₂) / (P₂Z₁T₁)

                                         = 200 ( 1 * 1* 360) / (493.45 *0.98*288.5)

                                         = 0.516 m^3/hr

Answer:

A. This option is true

B. This option is false

C. This option is true

Explanation:

A. Generally speaking, the greater percentage of cold, the recrystallization grain size would turn out to be smaller. Therefore this true.

B. A higher annealing temperature does not result in smaller recrystallization grain size. Therefore this is false.

C. As the percentage of cold work is greater, the recrystallization temperature would tend to be lower. Therefore this is true.

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

Complete Question:

Hans, a scheduler at Mangel-Wurzel Transport, gets a call from a regular customer needing to move 19.8m³ of rock and soil, which Hans knows from previous experience has an average density of 650/kgm³. Hans has available a dump truck with a capacity of 4m³ and a maximum safe load of 3700kg.  Required: calculate the number of trips the dump truck will have to make to haul the customer's load away.

Answer:

Mangel-Wurzel Transport

The number of trips that the dump truck will have to make to haul the customer's load away is:

= 5 trips.

Explanation:

a) Data and Calculations:

Volume of customer's load (rock and soil) = 19.8m³

Density of load = 650 kg/m³

Mass of load = Volume of load * Density of load

= 19.8m³ × 650 kg/m³

= 12,870 kg

The maximum safe load (mass) of the dump truck = 3,700 kg

Volume of the dump truck = 4m³  

Assuming the truck is to carry 4m³ of the load.

The mass of load that the 4m³ capacity truck can carry = 4m³ × 650kg/m³

= 2,600kg

Quick Check:

Mass = 2,600kg < 3,700 kg, satisfying required conditions.  

The number of trips that the truck would make to haul the customer's load away is, therefore, calculated as follows:

Number of trips = N

N = total volume of load/ volume per trip

N = 19.8/4

N = 4.95

N = 5 trips approx.

Answer:

a)  Δh = 2 cm,  b) Δh = 0.4 cm

Explanation:

Let's start by using Bernoulli's equation for the Pitot tube, we define two points 1 for the small entry point and point 2 for the larger diameter entry point.

            P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

Point 1 is called the stagnation point where the fluid velocity is reduced to zero (v₁ = 0), in general pitot tubes are used  in such a way that the height of point 2 of is the same of point 1

           y₁ = y₂

subtitute

           P₁ = P₂ + ½ ρ v₂²

           P₁ -P₂ = ½ ρ v²

where ρ is the density of fluid  

now we measure the pressure on the included beforehand as a pair of communicating tubes filled with mercury, we set our reference system at the point of the mercury bottom surface

           ΔP =ρ_{Hg} g h - ρ g h

           ΔP =  (ρ_{Hg} - ρ) g h

as the static pressure we can equalize the equations

          ΔP = P₁ - P₂

         (ρ_{Hg} - ρ) g h = ½ ρ v²

         v = [tex]\sqrt{\frac{2 (\rho_{Hg} - \rho) g}{\rho } } \ \sqrt{h}[/tex]

in this expression the densities are constant

        v = A  √h

       A =[tex]\sqrt{\frac{2(\rho_{Hg} - \rho ) g}{\rho } }[/tex]

They indicate the density of mercury rhohg = 13600 kg / m³, the density of dry air at 20ºC is rho air = 1.29 kg/m³

we look for the constant

        A = [tex]\sqrt{\frac{2( 13600 - 1.29) \ 9.8}{1.29} }[/tex]

        A = 454.55

we substitute

       v = 454.55 √h

to calculate the uncertainty or error of the velocity

         h = [tex]\frac{1}{454.55^2} \ v^2[/tex]

       Δh = [tex]\frac{dh}{dv}[/tex]   Δv

       [tex]\frac{\Delta h}{h } = 2 \ \frac{\Delta v}{v}[/tex]

Suppose we have a height reading of h = 20 cm = 0.20 m

a) uncertainty 2.5 m / s ( 0.05)

        [tex]\frac{\delta v}{v} = 0.05[/tex]

       [tex]\frac{\Delta h}{h}[/tex] = 2 0.05  

       Δh = 0.1 h

       Δh = 0.1  20 cm

       Δh = 2 cm

b) uncertainty 0.5 m / s ( Δv/v= 0.01)

        [tex]\frac{\Delta h}{h}[/tex] =  2 0.01

        Δh = 0.02 h

        Δh = 0.02 20

        Δh = 0.1 20 cm

        Δh = 0.4 cm = 4 mm

Answer: [tex]10.631\times 10^3\ N/m^2[/tex]

Explanation:

Given

Discharge is [tex]Q=12.5\ L[/tex]

Diameter of pipe [tex]d=150\ mm[/tex]

Distance between two ends of pipe [tex]L=800\ m[/tex]

friction factor [tex]f=0.008[/tex]

Average velocity is given by

[tex]\Rightarrow v_{avg}=\dfrac{12.5\times 10^{-3}}{\frac{\pi }{4}(0.15)^2}\\\\\Rightarrow v_{avg}=\dfrac{15.9134\times 10^{-3}}{2.25\times 10^{-2}}\\\\\Rightarrow v_{avg}=7.07\times 10^{-1}\\\Rightarrow v_{avg}=0.707\ m/s[/tex]

Pressure difference is given by

[tex]\Rightarrow \Delta P=f\ \dfrac{L}{d}\dfrac{\rho v_{avg}^2}{2}\\\\\Rightarrow \Delta P=0.008\times \dfrac{800}{0.15}\times \dfrac{997\times (0.707)^2}{2}\\\\\Rightarrow \Delta P=10,631.45\ N/m^2\\\Rightarrow \Delta P=10.631\ kPa[/tex]

Answer:

The right answer is "2.2099 m³".

Explanation:

Given:

Mass,

m = 5 kg

Temperature,

T = 35℃

or,

  = 35 + 273

Pressure,

P = 200 kPa

Gas constant,

R = 0.2870 kj/kgK

By using the ideal gas equation,

The volume will be:

⇒ [tex]PV=mRT[/tex]

or,

⇒    [tex]V=\frac{mRT}{P}[/tex]

By substituting the values, we get

          [tex]=\frac{5(0.2870)(35+273)}{200}[/tex]

          [tex]=\frac{441.98}{200}[/tex]

          [tex]=2.2099 \ m^3[/tex]

Answer:

first mark me as a brainleast

Answer:

The container should be made of lead and the liquid should be water.

Explanation:

Since the volume of the container of liquid after expansion is V = V₀(1 + βΔθ) where V = initial volume, β = coefficient of volume expansion, Δθ = temperature change.

So, the volume change V₂ - V₁ where V₁ = volume of liquid and V₂ = volume of container

For steel, V₂ = V₀(1 + β₂Δθ) and V₁ = V₀(1 + β₁Δθ)

So, ΔV = V₀(1 + β₂Δθ) - V₀(1 + β₁Δθ) = V₀[1 + β₂Δθ - 1 - β₁Δθ] = V₀[β₂Δθ - β₁Δθ]

Since we want a minimum value for ΔV and V₀ and Δθ are the same, we need ΔV/V₀Δθ = β₂ - β₁ to be a minimum

where β₂ = coefficient of volume expansion of liquid and β₁ = coefficient of volume expansion of container.

So, trying each combination, with  β₂ = 207 × 10⁻⁶ (C°)⁻¹] and β₁ = 36 × 10⁻⁶ (C°)⁻¹

β₂ - β₁ = 207 × 10⁻⁶ (C°)⁻¹ - 36 × 10⁻⁶ (C°)⁻¹ = 171 × 10⁻⁶ (C°)⁻¹

With  β₂ = 207  × 10⁻⁶ (C°)⁻¹] and β₁ = 87  × 10⁻⁶ (C°)⁻¹

β₂ - β₁ = 207 × 10⁻⁶ (C°)⁻¹ - 87 × 10⁻⁶ (C°)⁻¹ = 120 × 10⁻⁶ (C°)⁻¹

With  β₂ = 1120 × 10⁻⁶ (C°)⁻¹] and β₁ = 36  × 10⁻⁶ (C°)⁻¹

β₂ - β₁ = 1120 × 10⁻⁶ (C°)⁻¹ - 36 × 10⁻⁶ (C°)⁻¹ = 1084 × 10⁻⁶ (C°)⁻¹

With  β₂ = 1120 × 10⁻⁶ (C°)⁻¹] and β₁ = 87 ×  10⁻⁶ (C°)⁻¹

β₂ - β₁ = 207 × 10⁻⁶ (C°)⁻¹ - 87 × 10⁻⁶ (C°)⁻¹ = 1033 × 10⁻⁶ (C°)⁻¹

The combination that gives the lowest value for β₂ - β₁ is β₂ = 207 × 10⁻⁶ (C°)⁻¹] and β₁ = 87 × 10⁻⁶ (C°)⁻¹

Since  β₁ = 87 × 10⁻⁶ (C°)⁻¹ = coefficient of expansion for lead β₂ = 207 × 10⁻⁶ (C°)⁻¹]  = coefficient of expansion for water, the container should be made of lead and the liquid should be water.

Answer:

the current consumed is 3.3 A

Explanation:

Given;

resistance, R = 30 ohms

inductance, L = 200 mH

Voltage supply, V = 230 V

frequency of the coil, f = 50 Hz

impedance, Z = 69.6 Ohms

The current consumed is calculated as;

[tex]I = \frac{V}{Z} \\\\I = \frac{230}{69.6} \\\\I = 3.3 \ A[/tex]

Therefore, the current consumed is 3.3 A

answer

d just too the test

1.5 nautical miles per 1,000 feet

Answer:

53 sec / cycle

Explanation:

Displayed red time = 35 seconds

Start up lost time = 2 seconds

Yellow time = 4 seconds

Red time = 1 second

Total lost time L = 2n + r

L = lost time

n = number of phase

R = red time

35+2+4+4*1

= 45

L = 2x4+45

= 53 sec/cycle

The total lost time is typically calculated as 53 seconds per cycle

Answer:

A. The biceps muscle contracts while the triceps muscle relaxes. Only then can the forearm move up to lift the mug.

Explanation:

Skeletal muscle is also known as voluntary muscle and it can be defined as a type of muscle connected with the skeleton by tendons, so as to form a mechanical system that enables the movement of the limbs and other body parts with respect to another.

Generally, skeletal muscles are only found in vertebrates.

The muscles in the arm work to lift a mug of coffee to your lips when the biceps muscle contracts while the triceps muscle relaxes.

Basically, the only way the forearm move up to lift a mug for example, is through the contraction of the biceps muscle and the relaxation of the triceps muscle.

This ultimately implies that, the elbow joint is straightened or bent due to the actions of the biceps muscle and triceps muscle against each other.

Answer:A

Explanation:

got it right on plato

Answer:

the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

Explanation:

Given:

Diameter of the pipe = 100mm = 0.1m

Contraction ratio = 0.5

thus, diameter at the throat of venturimeter = 0.5×0.1m = 0.05m

The formula for discharge through a venturimeter is given as:

Where,

is the coefficient of discharge = 0.97 (given)

A₁ = Area of the pipe

A₁ =  

A₂ = Area at the throat

A₂ =  

g = acceleration due to gravity = 9.8m/s²

Now,

The gauge pressure at throat = Absolute pressure - The atmospheric pressure

⇒The gauge pressure at throat = 2 - 10.3 = -8.3 m (Atmosphric pressure = 10.3 m of water)

Thus, the pressure difference at the throat and the pipe = 3- (-8.3) = 11.3m

Substituting the values in the discharge formula we get

or

or

Q = 29.28 ×10⁻³ m³/s

Hence, the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

Hope This Helps :D

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

Answer:

a)  Vr = - a^2/r cosθ  + aß / r

    Vθ = 1/r [ -a^2/r * sinθ ]

b) attached below

Explanation:

potential function

Ø= a^2 /r  cosØ + aßlnr ----- ( 1 )

a = radius ,  ß = constant

a) Expressions for Vr and Vθ

Vr =  dØ / dr  ----- ( 2 )

hence expression : Vr = - a^2/r cosθ  + aß / r

Vθ = 1/r dØ / dθ ------ ( 3 )

back to equation 1

dØ / dr = - a^2/r sinθ + 0  --- ( 4 )

Resolving equations 3 and 4

Vθ = 1/r [ -a^2/r * sinθ ]

b) expression for stream function

attached below

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.

Answer:

Final mass of Argon=  2.46 kg

Explanation:

Initial mass of Argon gas ( M1 ) = 4 kg

P1 = 450 kPa

T1 = 30°C = 303 K

P2 = 200 kPa

k ( specific heat ratio of Argon ) = 1.667

assuming a reversible adiabatic process

Calculate the value of the M2

Applying ideal gas equation ( PV = mRT )

P₁V / P₂V = m₁ RT₁ / m₂ RT₂

hence : m2 = P₂T₁ / P₁T₂ * m₁

                   = (200 * 303 ) / (450 * 219 ) * 4

                   = 2.46 kg

Note: Calculation for T2 is attached below