plsssssss help me here​

Answer:

There are several different types of charts and graphs. The four most common are probably line graphs, bar graphs and histograms, pie charts, and Cartesian graphs. They are generally used for, and are best for, quite different things. ... Pie charts to show you how a whole is divided into different parts.

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Explanation:

This relationship, known as Faraday's law of induction (to distinguish it from his laws of electrolysis), states that the magnitude of the emf induced in a circuit is proportional to the rate of change of the magnetic flux that cuts across the circuit.

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%

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

Answer:

7.9 billion people

Explanation:

Answer:

a. Rockwell              3. hardness

b. Instron                 2. stress vs strain

c. Charpy                 1. impact strength

d. Fatigue                4. Endurance Limit

e. Brinell                  3. hardness

f. Izod                      1. impact strength

Explanation:

Izod and Charpy are the impact strength testing procedure of a material in which a heavy hammer is attached to an arm is released to impact on the test specimen. In Izod test the specimen with v-notch is held vertical with the notch facing outward while in Charpy test the specimen is supported horizontally with notch facing inward to the impacting hammer.

Instron testing system does universal testing of the material which gradually applies the load recording all the stresses and the corresponding strains until the material fails.

Fatigue is the property of a material due to which it fails under the repeated cyclic loading by the initiation and propagation of cracks. The property of a material resist failure subjected to infinite number of repeated cyclic loads below a certain stress limit.

Rockwell and Brinell are the hardness testing methods. In Rockwell test an intender ball is firstly pressed against the specimen using minor load for a certain time and then a major load is pressed against it for a certain time. After the intender is removed the depth of impression on the surface is measured while in case of Brinell hardness we apply only one load against the intender ball for a certain time and after its removal the radius of impression is measured.

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]

Answer:

I can't understand this language .

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

Answer:

1.c

2.c

3.c

Explanation:

James is a  pilot, whistle. He is wearing a flight suit. Paris is the palace where does he fly to. Hence, option C, C, and C are correct.

When flying an aircraft, such as a military aircraft, a glider, or a helicopter, one must wear a full-body suit called a flight suit. These outfits are typically meant to keep the user warm and are also functional (they have many of pockets) (including fire ). In most cases, it looks like a jumpsuit.

The G suit, sometimes known as a "anti-G suit," is a one-piece jumpsuit that shields a pilot from the pressure of G forces pressing down on him and causing discomfort or unconsciousness.

The traditional attire for pilots of military and commercial aircraft, helicopters, and even gliders is flight suits or flyers coveralls. In areas where there is a risk of fire, ground personnel—including aircrews—often wear flight suits as well.

Thus, option C, C, and C are correct.

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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:

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Explanation:

j

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

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.

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:

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;

}

}

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

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.

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:

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.

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:

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.

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:

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]      

Answer:

[tex]h_{1} = h_2} = 293.45 KJ/kg[/tex].

The quality of the refrigerant exiting the expansion valve is

[tex]x_{2}=0.193596[/tex].

Explanation:

Fluid given Ammonia.

Inlet 1:-

Temperature [tex]T_{1}[/tex] = [tex]24^{o} C[/tex].

Pressure [tex]P_{1}[/tex] = 10 bar.

Exit 2:-

Pressure [tex]P_{2}[/tex] = 1 bar.

Solution:-

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.

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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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.

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]

[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

[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

[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