A sign erected on uneven ground is guyed by cables EF and EG. If the force exerted by cable EF at E is 46 lb, determine the moment of that force about the line joining points A and D.

Defining the radius of area throat to inlet we would have that the proportional relationship would be [tex]\frac{A_2}{A_1}.[/tex]

This equation or relationship is obtained from continuity where

[tex]A_1V_1 = A_2V_2[/tex]

[tex]\frac{V_2}{V_1} = \frac{A_2}{A_1}= 0.8[/tex]

Now applying the Bernoulli equation between inlet and throat section we have,

[tex]\frac{p_1}{\rho g}+ \frac{v_1^2}{2g}+z_1=\frac{p_2}{\rho g}+ \frac{v_2^2}{2g}+z_2[/tex]

Here,

[tex]z_1 = z_2[/tex]

Then for a Venturi duct, the velocity of the airplane [tex]V_1[/tex] will be

[tex]V = \sqrt{\frac{2(p_1-p_2)}{\rho[(\frac{A_1}{A_2})^2-1]}}[/tex]

Our values are,

[tex]\frac{A_2}{A_1} = 0.8[/tex]

[tex]\rho = 0.002377slug/ft^3[/tex]

[tex]p_1 = 2116lb/ft^2[/tex]

[tex]p_2 = 2100lb/ft^2[/tex]

Replacing,

[tex]V= \sqrt{\frac{2(2116-2100)}{(0.002377)[(\frac{1}{0.8})^2-1]}}[/tex]

[tex]V = 154.7ft/s[/tex]

Therefore the velocity of the airplane is 154.7ft/s

Answer:

The engine's thermal efficiency is 0.32

Explanation:

Thermal efficiency = work done ÷ quantity of heat supplied

Work done = 210 J

Quantity of heat supplied = work done + waste heat = 210 + 440 = 650 J

Thermal efficiency = 210 ÷ 650 = 0.32

Answer:

T₂ =93.77  °C

Explanation:

Initial temperature ,T₁ =27°C= 273 +27 = 300 K

We know that

Absolute pressure = Gauge pressure + Atmospheric pressure

Initial pressure ,P₁ = 300+1=301 kPa

Final pressure  ,P₂= 367+1 = 368  kPa

Lets take  temperature=T₂

We know that ,If the volume of the gas is constant ,then we can say that

[tex]\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}[/tex]

[tex]{T_2}=T_1\times \dfrac{P_2}{P_1}[/tex]

Now by putting the values in the above equation we get

[tex]{T_2}=300\times \dfrac{368}{301}\ K[/tex]

The temperature in  °C

T₂ = 366.77 - 273  °C

T₂ =93.77  °C

Answer: The monthly payment before the buydown is $71.3

The monthly payment after the buydown is $68.9

Explanation: The payment is compounding so we use compound interest;

A= P[1+(r/n)^nt]

Where;

A= Compounded amount

P = principal

r= interest rate per payment

n= number of payment per period

t= number of period.

NOTE: from our questions, the period is yearly and the payment is monthly. Therefore;

number of payment per period (n) is 12

number of payment period (t) is 10

P=$8000, r= 0.667% or 0.333%

FIND MONTHLY PAYMENT BEFORE BUYDOWN:

Step 1: find the Compounded amount to pay.

A= $8000[1+(0.00667÷12)^(12×10)]=

$8551.64 this is the total amount he has to pay for a period of 10years

Step 2: How much does he has to pay monthly for a period of 10year;

Therefore his payment will be for 120 months

$8551.64÷120= $71.3 monthly

FIND MONTHLY PAYMENT AFTER BUYDOWN:

Step 1: find the compounded amount to pay.

A= 8000[1+(0.00333÷12)^(12×10)=

$8270.85 this is the total amount he has to pay for a period of 10years

Step2: How much does he has to pay monthly for a period of 10year;

Therefore his payment will be for 120 months;

$8270.85÷120= $68.9 monthly

To calculate the monthly payment for a $8,000 loan at 8% interest over 10 years, the loan payment formula is applied with a monthly interest rate of 0.667%. After applying a 4% interest buydown, reducing the rate to 4%, the same formula is used with the new monthly rate of 0.333% to find the reduced monthly payment. Both calculations highlight the financial benefits of an interest buydown.

Calculating Monthly Payments Before and After Interest Buydown

To calculate the monthly payment for a loan of $8,000 at an 8% base rate over 10 years with monthly compounding, we need to use the formula for an installment loan which includes the principal and the interest. Similarly, we'll calculate the reduced payment after a 4% interest buydown to 4%.

The monthly payment before the buydown, using a monthly interest rate of 0.667%, can be determined using the formula:

M = P * (r(1+r)^n) / ((1+r)^n - 1)

Where:

Plugging in the values, we calculate the monthly payment before the buydown.

The monthly payment after the buydown, with a reduced monthly interest rate of 0.333%, is calculated in a similar way.

In both cases, the formula provides us with the monthly payment amounts, showing the impact of the interest buydown on the monthly payments.

Answer:

9.248 < [tex]\mu[/tex] < 9.253 mm

Explanation:

Given data:

standard deviation  = 0.021 mm

sample mean = 9.251 mm

total sample = 409 m

confidence interval level = 95%

mean diameter of entire batch  [tex]\mu  = \bar X \pm z \frac{\sigma}{\sqrt{n}}[/tex]

where

[tex] \bar X [/tex] - sample mean = 9.251 mm

z = critical value

plugginf all value in the above relation to get the mean diameter

[tex]\mu = 9.251 \pm 1.96 \times \frac{0.021}{\sqrt{409}}[/tex]

9.248 < [tex]\mu[/tex] < 9.253 mm

Answer:

v_2 = 972 π inch^3

Explanation:

Given data:

volume of air V_1 = 288π inc^3

radius of ball at 1 beach = r_1

radius of ball at 2nd beach = r_1 + 3

we know that [tex]V_1 = 288 \pi[/tex]

we know that volume is given as [tex]v_1 =  \frac{4}{3} \pi r_1^3[/tex]

so equating both side of volume we get

[tex]288\pi =  \frac{4}{3} \pi r_1^3[/tex]

r_1 = 6 inch

therefore r_2 = 9 inch

volume of air at 2nd beach [tex]v_2 =  \frac{4}{3} \pi 9^3[/tex]

v_2 = 972 π inch^3

Answer:

2.0978 g/cm^3

Explanation:

Given data:

[tex]p=1.91 g/cm^{3}[/tex]

w = 9.5 %

[tex]G_{s}[/tex]= 2.70

e =?

[tex]S_{r}[/tex] =?

solution:

[tex]w=\frac{mass of water}{mass of solid} =\frac{m_{w} }{m_{s} }[/tex]

[tex]e=\frac{volume of solid}{volume of solid} =\frac{V_{s} }{V_{s} }[/tex]

assume total volume [tex]V_{total}[/tex]=1 [tex]cm^{3}[/tex]=[tex]V_{w} +V_{s} +V_{air}[/tex]

[tex]p=\frac{m_{total} }{V_{total} }[/tex]

[tex]m_{total} =1.91 g\\m_{total} =m_{w} +m_{s}\\w=\frac{m_{w} }{m_{s}}\\ 0.095.m_{s}=m_{w}\\1.91=m_{s}+0.095.m_{s}\\m_{s}=1.744 g\\m_{w}=0.166 g\\[/tex]

[tex]p_{w} =\frac{m_{w} }{V_{w} } ==> V\\G_{S}=\frac{m_{s} }{V_{s}.p_{w}}\\2.70=\frac{1.744}{V_{s}.1 } \\\V_{s}=0.646 cm^{3} \\V_{v}=V_{T}-V_{S}=0.354 cm^{3} \\e=\frac{V_{v}}{V_{S}} =\frac{0.354}{0.646}=0.55=55 percent\\[/tex]

[tex]S_{r}=\frac{w*Gs}{e} =0.4666=46.6 percent\\if S_{r}=1 \\ w=??\\S_{r}=\frac{w*Gs}{e} \\\\w=0.204=20.4 percent\\[/tex]

now find [tex]p[/tex]

[tex]p=\frac{m_{t}(=m_{w} +m_{s} )}{V_{t} }[/tex]

[tex]p[/tex]=2.0978 g/cm^3

Answer:

Problem 1 (10 points) In the first homework you were instructed to design the mechanical components of an oscillating compact disc reader. Since you did such a good job in your design, the company decided to work with you in their latest Blue-ray readers, as well. However, this time the task is that once the user hits eject button, the motor that spins the disc slows down from 2000 rpm to 300 rpm and at 300 rpm a passive torsional spring-damper mechanism engages to decelerate and stop the disc. Here, your task is to design this spring-damper system such that the disc comes to rest without any oscillations. The rotational inertia of the disc (J) is 2.5 x 10-5kg m² and the torsional spring constant (k) is 5 × 10¬³NM. Calculate the critical damping coefficient cc for the system. choice of the damper, bear in mind that a good engineer stays at least a factor of In your 2 away from the danger zone (i.e., oscillations in this case). Use the Runge Kutta method to simulate the time dependent angular position of the disc, using the value of damping coefficient (c) that calculated. you Figure 1: Blue-ray disc and torsional spring-damper system.

a) The water content is 85.3%.

b) The void ratio is approximately 2.31.

c) The degree of saturation is 99.6%.

How do we determine the water content?

a) We shall calculate the water content (w) using the formula:

[tex]\frac{M_{w} }{Ms} * 100[/tex]

where:

[tex]M_{w}[/tex] = mass of water

[tex]M_{s}[/tex] = mass of solids (oven-dried mass)

Given:

[tex]M_{w}[/tex] = [tex]417g -225g = 192g[/tex]

[tex]M_{s}[/tex] = [tex]225g[/tex]

We shall now calculate water content (w):

[tex]w = (192/225) * 100[/tex]

[tex]w = 0.853 * 100[/tex]

[tex]w = 85.3[/tex]%

Thus,  the water content is 85.3%.

b) We shall compute the Void ratio (e) by using the formula:

[tex]e = \frac{V_{v} }{V_{s} }[/tex]

where:

[tex]V_{v} =[/tex] volume of voids

[tex]V_{s} =[/tex] volume of solids

Let us find [tex]V_{s}[/tex] first:

[tex]V_{s} =[/tex] [tex]M_{s} /[/tex]Specific gravity * Density of water

[tex]V_{s} = \frac{225g}{2.70 * 1g/cm^{3} }[/tex]

[tex]V_{s} = \frac{225g}{2.70g/cm^{3} }[/tex]

[tex]V_{s} = 83.33cm^{3}[/tex]

[tex]V_{v} = V_{t} - V_{s}[/tex]

[tex]V_{v} = 276cm^{3} - 83.33cm^{3}[/tex]

= 192.67cm³

We calculate e (Void ratio):

e = [tex]192.67/83.33[/tex]

[tex]e = 2.31.[/tex]

Hence, the void ratio is ≈ 2.31.

c) We can find the degree of saturation (S) using the formula:

[tex]S = \frac{V_{w} }{V_{v} }[/tex]

Given:

[tex]V_{w}[/tex] = 192cm³

And we have [tex]V_{v}[/tex] = 192.67cm³

[tex]S = (192/192.67) * 100[/tex]

[tex]S = 0.996 * 100[/tex]

S = 99.6%

So, the degree of saturation is 99.6%.

Answer:

Lets check each statement for the errors.

Explanation:

System.out.println("Predictions are hard.");

This statement has no syntax errors. When this statement is executed the following line will be displayed:

Predictions are hard.

System.out.print("Especially ');

This statement is missing the closing quotation mark. A single quotation mark is placed instead of double quotation mark in the statement.

The following error message will be displayed when this program statement will be compiled:

Main.java:15: error: unclosed string literal

String literals use double quotes. So to correct this syntax error, the statement should be changed as follows:

System.out.print("Especially");

The output of this corrected line is as following:

Especially

System.out.println("about the future.").

In this line a period . is placed at the end of the statement instead of a semicolon ; but according to the syntax rules statements should end in semicolons.

The error message displayed when this line is compiled is as following:

Main.java:15: error: ; expected

Main.java:15: error: not a statement

So in order to correct this syntax error the statement should be changed as following:

System.out.println("about the future.");    

The output of this corrected line is as following:

about the future

System.out.println("Num is: " - userName);

There is a syntax error in this statement because of - symbol used instead of +

+ symbol is used to join together a variable and a value a variable and another variable in a single print statement.

The error message displayed when this line is compiled is as following:

Main.java:13: error: bad operand types for binary operator '-'

So in order to correct this syntax error the statement should be changed as following:

System.out.println("Num is: " + userName);

This line will print two things one is the string Num is and the other is the value stored in userName variable.

So let userName= 5 then the output of this corrected line is as following:

Num is: 5

Answer:

A) The trajectory traced by the drifting marker form a pathline. The pathline is the trajectory than a single particle does in the fluid (like in this case) along with all its travel.

b) The line connecting these markers released at the same point in space in quick succession is a Streakline. The streaklines are the lines formed by the loci of points in the fluid that have continuously pass throw particulars points in space in the past. The main difference between the pathline and the streakline is that the first one is not affected by flow changes over time, while the second one is. The Pathline is a movie that shows the travel of a single particle in time. The streakline is movie that shows how the particles change their path along time.

In the gif attached to this answer, you can see in red the pathline of particle, in blue the streakline of several particles released at the same point in space in quick succession and in grey the streamline.

Answer:

a) T_5 = 782.8 K

b)  W_cyc = 108.04 KJ/kg

c) n_th = 22.47 %

d) X_dest = 289.924 KJ/kg , X_exhaust = 126.6768 KJ/kg

Explanation:

Given:

- P_2 / P_1 = 7

- T_4 = 1150 K

- T_1 = 310 K

- n_s,comp = 0.75

- n_s,turb = 0.82

- R_air = 0.287 KJ/kg

Find:

- T_5 - Temperature of air at turbine exit ?

- W_cycle ?

- n_th ?

- X_dest , X_exhaust ?

Solution:

Assumptions:

1) The cycle operates at steady-state.  

2) Air is the working fluid and it behaves as an ideal gas.

3) The Brayton Cycle is modeled as as a closed cycle.

4) The combustor is replaced by a HEX. (External Combustion)

5) The compressor and turbine are not internally reversible.

6) Changes in kinetic and potential energies are negligible.

7) Air has variable specific heats.

8) The compressor and turbine are adiabatic.

Analysis:

- The efficiency of turbine is given by:

                            n_s,turb = (H_4 - H_5) / (H_4 - H_5,s)

- For H_4 and S_ 4 we have T_4 = 1150 K, use the ideal gas air property table:

          T_4 = 1150 K  -------------> S_4 = 3.129 KJ/kgK   ,  H_4 = 1219.25 KJ/kg

- For H_5s,  the enthalpy of the effluent from a hypothetical isentropic turbine. We can do this because  we know the values of two intensive variables: P_5 and S_5 = S_4. The key to using this information is the 2nd Gibbs equation:

          S_5,s,o = S_4,s,o + R_air*Ln(P_5 / P_4)

          S_5,s,o = 3.129 + 0.287*Ln(1 / 7) = 2.57054 KJ/kgK

- Now use the value S_5,s,o and ideal gas air property table and evaluate:

           S_5,s,o = 2.57054 KJ/kgK ---------> T_5,s = 698.6 K ,

                                                                      H_5s = 711.72 KJ/kg

- Now use the efficiency relation for turbine:

            H_5 =   H_4 - n_s,turb*(H_4 - H_5,s)

            H_5 = 1219.25 - 0.82*(1219.25-711.72)

            H_5 = 803.08 KJ/kg

- Using H_5 and ideal gas air property table and evaluate:

         H_5 = 803.08 KJ/kg -----------> T_5 = 782.8 K , S_5 = 2.6940 KJ/kg-K

- In order to determine the specific shaft work for the cycle, we need to determine the specific shaft work for the compressor  and for the turbine

             W_cyc = W_comp + W_turb

             W_cyc = H_1 - H_2 + H_4 - H_5

- The efficiency of compressor is given by:

                            n_s,comp = (H_1 - H_2,s) / (H_1 - H_2)

- For H_1 and S_ 1 we have T_1 = 310 K, use the ideal gas air property table:

          T_1 = 310 K  -------------> S_1 = 1.73498 KJ/kgK , H_1 = 310.24 KJ/kg

- For H_2s,  the enthalpy of the effluent from a hypothetical isentropic compressor. We can do this because  we know the values of two intensive variables: P_2 and S_2 = S_1. The key to using this information is the 2nd Gibbs equation:

          S_2,s,o = S_1,s,o + R_air*Ln(P_2 / P_1)

          S_2,s,o = 1.73498 + 0.287*Ln(7) = 2.29343 KJ/kgK

- Now use the value S_2,s,o and ideal gas air property table and evaluate:

           S_2,s,o = 2.29343 KJ/kgK ---------> T_2,s = 537.1 K ,

                                                                      H_2s = 541.34 KJ/kg

- Now use the efficiency relation for compressor:

            H_2 =   H_1 - (H_1 - H_2,s)/n_comp

            H_2 = 310.24 - (310.24-541.34)/0.72

            H_2 = 618.37 KJ/kg

Hence,

The work out for the cycle is:

            W_cyc = 310.24 - 618.37 + 1219.25 - 803.08

            W_cyc = 108.04 KJ/kg

- The thermal efficiency of a cycle is:

            n_th = W_cyc / Q_H

            Q_H = H_4 - H_3

- The effectiveness of re-generator is e:

            e = (H_3 - H_2) / (H_5 - H_2)

            H_3 = (H_5 - H_2)*e + H_2

            H_3 = (803.08 - 618.37)*0.7 + 618.37 = 738.43 KJ/kg

Hence,

            Q_H = 1219.25 - 738.43 = 480.81 KJ/kg

Finally,

             n_th = 108.04 / 480.81 = 22.47 %

- The amount of heat loss is given by:

              Q_L = H_6 - H_1

              H_6 = H_5 + H_2 - H_3 = 803.08 + 618.37 - 738.43 = 682.97 KJ/kg

              Q_L = 682.97 - 310.24 = 372.73 KJ/kg

- The amount of exergy destroyed for whole cycle:

              X_dest = T_L * ( Q_L / T_L - Q_H / T_H)

              X_dest = 310 * (372.73 / 310 - 480.81 / 1800)

              X_dest = 289.924 KJ/kg

- The amount of exergy of exhaust gasses:

              X_exhaust = H_6 - H_0 - T_L*(S_6 - S_o )

              X_exhaust = 682.97 - 310.24 - 310*(2.52861 - 1.73489 )

              X_exhaust = 126.6768 KJ/kg

Answer:

(a) the pressure and temperature at the end of the heat addition process is 1733.79 K and 4392.26 Kpa respectively

(b) the net work output is 423.54 KJ/Kg

(c) the thermal efficiency is 56.5%

(d) the mean effective pressure for the cycle cannot be determined without initial volume of the process

Explanation:

Assumptions:

The properties of air at room temperature;

Cp = 1.005 KJ/kg.K, Cv = 0.718KJ/kg.K, R = 0.287KJ/kg.K and K = 1.4

Part a:

For isentropic compression:

[tex]T_2=T_1[\frac{V_1}{V_2}]^{K-1}[/tex]

Where;

T₁ = (27+273)K =300K

V₁/V₂ = 8

[tex]T_2=300[8]^{1.4-1} = 300[8]^{0.4} = 689.22K[/tex]

Based on the assumption above;

Q₁ₙ = U₃ -U₂

For an ideal gas with constant specific heats, the change in internal energy in terms of the change in temperature is shown below

U₃ -U₂ = Cv(T₃-T₂)

Thus; Q₁ₙ = Cv(T₃-T₂)

T₃ = (Q₁ₙ/Cv) + T₂

T₃ = (750/0.718) + 689.22 K = 1733.79 K

From general gas equation, we find the second stage pressure

[tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}[/tex]

[tex]P_2 =P_1[\frac{T_2}{T_1}][\frac{V_1}{V_2}] = 95 Kpa[\frac{689.22}{300}](8)[/tex]

P₂ = 1746.024Kpa

To obtain the pressure at stage 3

[tex]P_3 =P_2[\frac{T_3}{T_2}][\frac{V_2}{V_3}] = 1746.024 Kpa[\frac{1733.79}{689.22}](1)[/tex]

P₃ = 4392.26 Kpa

Part b:

To obtain net work output we consider overall energy balance on the cycle

[tex]Q_{out} = U_4-U_1 = C_v(T_4-T_1)[/tex]

For is isentropic  expansion

[tex]T_4=T_3[\frac{V_3}{V_4}]^{K-1} = 1733.79 [\frac{1}{8}]^{0.4} = 754.68K[/tex]

[tex]Q_{out} = 0.718(754.68-300) = 326.46 KJ/Kg[/tex]

To solve for net work output:

[tex]Q_{net} = Q{in}- Q_{out}[/tex] = (750 - 326.46) KJ/Kg = 423.54 KJ/Kg

part c:

To calculate the thermal efficiency, we use net work output per input work

η = 423.54/750

η =  0.565 = 56.5%

part d:

Mean effective pressure for the cycle (MEP)

[tex]MEP = \frac{Q_{net}}{V_1-V_2}[/tex] = [tex]\frac{Q_{net}}{V_1(1-\frac{1}{r})} = \frac{423.54}{V_1(1-\frac{1}{8}) }[/tex]

MEP = [tex]\frac{Q_{net}}{V_1(1-\frac{1}{r})} = \frac{423.54}{V_1(0.875)}[/tex]

Thus mean effective pressure cannot be determined without initial volume of the process.

Answer:

Option B

Explanation:

An operational amplifier usually has a high open loop gain of around 10^5 which allows a wide range get of feed back levels in order to achieve the desired performance so therefore a low open loop gain reduces the range feed back level thereby reducing the performance which can cause errors in the output voltage.

Answer:

d. All of the above would require an EIS.

Explanation:

A document prepared with the aim of describing the impacts of suggested operations on the environment is an Environmental Impact Statement (EIS). There was a mistake. An Environmental Impact Statement (EIS) is therefore a report describing the environmental effects resulting from a current action. All of the activities above would have an effect on the environment and therefore must fill an EIS

Answer:

When an additional layer of insulation is applied to a cylindrical pipe or a spherical shell, the insulation layer works to increase its conduction resistance but at the same time, lowers the convection resistance of the surface.

Explanation:

Generally, when more insulation is added to a wall, the resulting effect is that heat transfer decreases. With increasing thickness of the insulation, the heat transfer rate becomes lesser. This is because the thermal resistance of the wall becomes more with the added insulation, wherein the heat transfer area and convection resistance aren't affected.

However, a different scenario occurs when an additional layer of insulation is applied to a cylindrical pipe or spherical shell. The conduction resistance increases, but the surface convection resistance decreases since the outer surface area for convection also increases and does not remain constant.

Overall, the heat transfer from the cylindrical pipe or spherical shell may increase or decrease, which depends on the effect that dominates.

Answer:

[tex]a=5.908\ \text{m/s$^2$}[/tex]

Explanation:

To get the magnitude of the acceleration you need to find the change in velocity across the nozzle. To do that you can use the continuity equation which stated that the flow rate is the same in every point in the flow. So in this case you define the points at the nozzle inlet and at the nozzle outlet:

[tex]q_v_1=q_v_2\\A_1v_1=A_2v_2\\[/tex]

To get the velocity at point 1 you can just divide the flow rate by the surface:

[tex]v_1=\cfrac{q_v1}{A_1}=0.207\ \text{m/s}[/tex]

Now the expression for the velocity at point 2 is:

[tex]v_2=\cfrac{A_1}{A_2}v_1=\cfrac{d_1^2}{d_2^2}v_1=1.294\ \text{m/s}[/tex]

Now if you assume linear change of velocity across the nozzle you can use the following expression for the acceleration:

[tex]a=\cfrac{2v_1^2}{L}\left(\cfrac{2x}{L}+1\right)[/tex]

where L is your nozzle length (20 cm).

This equation depends on the position of your particle moving down the centerline of the nozzle. I sketched it:

The electronics can be operated for approximately 25.19 hours from the fully charged battery, which stores 1.26 kWh of energy. The cost of the energy per kilowatt hour for the battery, given its lifecycle, is around $0.20.

The electronics aboard a sailboat consume 50 watts from a 12.6-V source, and the storage battery used is rated for 12.6 V and 100 ampere hours (Ah). To calculate the number of hours the electronics can be operated without recharging, we use the power consumption and the voltage to find the current drawn by the electronics:

I = P/V = 50W / 12.6V = 3.968 Ah

Thus, the electronics draw approximately 3.97 Ah from the battery. Since the battery has a capacity of 100 Ah, the operating time before a recharge is needed can be calculated as:

operating time = battery capacity / current draw = 100 Ah / 3.97 Ah ≈ 25.19 hours

To find how much energy in kilowatt-hours is stored in the battery:

Energy = Voltage  imes Capacity = 12.6V  imes 100Ah = 1260 Wh = 1.26 kWh

Considering the battery's life of 300 charge-discharge cycles and cost of $75, the cost per kilowatt hour can be calculated as follows:

cost per kWh = total cost / (energy storage  imes number of cycles) = $75 / (1.26 kWh  imes 300 cycles) ≈ $0.20 per kWh

Answer:

Relative density = 0.545

Degree of saturation = 24.77%

Explanation:

Data provided in the question:

Water content, w = 5%

Bulk unit weight = 18.0 kN/m³

Void ratio in the densest state, [tex]e_{min}[/tex] = 0.51

Void ratio in the loosest state, [tex]e_{max}[/tex] = 0.87

Now,

Dry density, [tex]\gamma_d=\frac{\gamma_t}{1+w}[/tex]

[tex]=\frac{18}{1+0.05}[/tex]

= 17.14 kN/m³

Also,

[tex]\gamma_d=\frac{G\gamma_w}{1+e}[/tex]

here, G = Specific gravity = 2.7 for sand

[tex]17.14=\frac{2.7\times9.81}{1+e}[/tex]

or

e = 0.545

Relative density = [tex]\frac{e_{max}-e}{e_{max}-e_{min}}[/tex]

= [tex]\frac{0.87-0.545}{0.87-0.51}[/tex]

= 0.902

Also,

Se = wG

here,

S is the degree of saturation

therefore,

S(0.545) = (0.05)()2.7

or

S = 0.2477

or

S = 0.2477 × 100% = 24.77%

Answer:

Advantage of satellite communication are as follow

- it help in mobile and wireless communications

- it can covers the wide area in single time

- installation of satellite communication is easy

- it can be used for any form of communication i.e. video, audio, etc.

- service can be available in uniformly

Explanation:

Advantage of satellite communication are as follow

- it helps in mobile and wireless communications

- it can cover the wide-area in a single time

- installation of satellite communication is easy

- it can be used for any form of communication i.e. video, audio, etc.

- service can be available in uniformly

Distance insensitive communication system is describe in terms of limitation      discovered area. Example of this is mobile and satellite. The mobile discover area is less than satellite.  

To determine the minimum required diameter of an anchor rod, we must calculate the allowable tensile stress using the factor of safety and the ultimate failure stress, then solve the area formula for a circular cross-section for the rod's diameter.

To calculate the minimum required diameter of an anchor rod designed using Allowable Stress Design and considering a factor of safety (F.S.), we must ensure that the designed stress ( au_{allow}) does not exceed the allowable stress. The allowable stress is derived from the material's ultimate (failure) stress ( au_{fail}) divided by the factor of safety (F.S.). Given the allowable load (F_{allow}) the rod must support and the factor of safety (F.S.), the failure load (F_{fail}) is F_{fail}= F_{allow} \times F.S.

Since  au_{allow} = F_{allow} / A, rearranging to solve for the cross-sectional area (A) needed gives us A = F_{allow} /  au_{allow}. And for a circular cross-section rod, A = \pi (d/2)^{2}, where d is the diameter of the rod.

With the failure stress given as  au_{fail} = 60 MPa and the factor of safety F.S. = 1.6, the allowable tensile stress ( au_{allow}) is  au_{allow} =  au_{fail} / F.S. = 60 MPa / 1.6.

Using the formula A = \pi (d/2)^{2} and the given load F_{allow} = 11 kN, we can solve for d. First, we convert the load to Newtons (since 1 kN = 1000 N), then substitute the values of F_{allow} and  au_{allow} into the area equation and solve for the diameter (d).

Upon calculating d, we obtain the minimum required diameter of the rod that will ensure it supports the given load with the required factor of safety.

Factor of safety: The factor of safety is defined as the ratio of the load that causes failure to the load the member can safely support. In this case, the factor of safety for tension failure is given as 1.6. A factor of safety larger than 1 ensures structural stability.

Calculating required diameter: Using the allowable stress design approach, the minimum required diameter of the rod can be calculated to ensure it can support the load safely without failing in tension. Given the material failure stress and the factor of safety, the required diameter can be determined based on uniform stress distribution.

Engineering analysis: Engineers use factors of safety and stress calculations to design structural elements that can safely support loads. Understanding stress, strain, and material properties are essential in ensuring the structural integrity and safety of a design.

In this exercise we have to use electronic knowledge to calculate the voltage value from the resistance. In this way we can conclude that:

So from the data given in the text, we have that:

So with the formula given below, we can calculate what is being asked by it:

[tex]Peak \ to \ peak \ ripple = I (load)/(f)(c)[/tex]

They are using the data previously informed and putting it in the given formula, we would be with:

[tex]I (load) = load \ current = 30/600 = 0.05 A\\Peak \ to \ peak \ ripple = 0.05/6\\= 8.33mV[/tex]

The average Dc something produced power for a complete wave exist double :

[tex]Vd_c= 0.637 * 30\\Vd_c =19.11V[/tex]

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Answer:

attached below

Explanation:

The concept of equilibrium which is useful when dealing with forces is states that a body on which exactly balanced forces act has no net force acting on it and the body is said to be in equilibrium

The weight of the suspended block E is approximately 18.33-lb

The reason why the value of the weight is correct is as follows:

The given parameters are;

The length of the chord = 4-ft.

The weight of the block D = 10-lb

The horizontal distance between pin A and C = 1 ft.

The length of the segment of the string s  when the system is in equilibrium = 1.5 ft.

Required:

To determine the weight of the suspended block E, if the system is in equilibrium

Solution:

The tension in the cord, T = The weight of block D = 10-lb

By equilibrium of forces, we have;

E = 2 × T × cos(θ)

Where:

E = The weight of the block located at E

θ = The angle between segment CB and the vertical

[tex]sin(\theta) = \dfrac{Opposite}{Hypotenuse}[/tex]

The length of the opposite side to θ = 1 ft./2 = 0.5 ft.

The length of the hypotenuse side = [tex]\dfrac{4 - 1.5}{2} = 1.25[/tex]

Therefore;

[tex]sin(\theta) = \dfrac{0.5}{1.25} = 0.4[/tex]

θ ≈ 23.58°

E = 2 × 10 × cos(23.58°) ≈ 18.33 lb.

The weight of the suspended block E ≈ 18.33 lb.

Learn more about the equilibrium of forces here:

Answer:

V2 = final volume = 8.3m^3

Explanation:

Given P1 = 445 kPa, V1 = 2.6 m^3, P2 = 140 kPa

From PV = constant; P1V1 =P2V2 , where V2 = final volume

V2 = P1V1/P2

Substituting in the equation ;

V2 = 445 x 2.6 / 140

V2 = final volume = 8.3m^3

The question involves estimating the rate of ammonia diffusion through a stagnant air layer, applying Fick's first law of diffusion. A precise numerical answer cannot be provided without specific concentration values or boundary conditions. Understanding diffusion is crucial in many scientific and industrial applications.

The scenario provided involves the diffusion of ammonia through a stagnant layer of air, under specific environmental conditions. To estimate the rate of diffusion, Fick's first law of diffusion can be applied, which states that the diffusion flux is proportional to the concentration gradient across the stagnant layer. Given the concentration of ammonia at one boundary is 50% and negligible at the other, with a layer thickness of 1 mm (0.001 m), and a diffusivity of 1.8 x 10-5 m2/s, the rate of diffusion can be calculated.

Unfortunately, without the specific concentration values or boundary conditions beyond percentage, a numerical calculation cannot be accurately provided in this response. Generally, the rate of diffusion would be determined by multiplying the diffusivity by the concentration gradient across the layer and inversely proportional to the thickness of the layer.

It is essential in chemical engineering and various scientific fields to understand how substances move through mediums, whether in industrial applications, environmental science, or physiology. The principles governing diffusion are critical for designing systems for the separation, treatment, or transfer of materials.

Answer:

γ[tex]_{xy}[/tex] =0.01, P=248 kN

Explanation:

Given Data:

displacement = 2mm ;

height = 200mm ;

l = 400mm ;

w = 100 ;

G = 620 MPa = 620 N//mm²;    1MPa = 1N//mm²

a. Average Shear Strain:

The average shear strain can be determined by dividing the total displacement of plate by height

γ[tex]_{xy}[/tex] = displacement / total height

     = 2/200 = 0.01

b. Force P on upper plate:

Now, as we know that force per unit area equals to stress

τ = P/A

Also,  τ = Gγ[tex]_{xy}[/tex]

By comapring both equations, we get

P/A = Gγ[tex]_{xy}[/tex]   ------------ eq(1)

First we need to calculate total area,

A = l*w = 400 * 100= 4*10^4mm²

By putting the values in equation 1, we get

P/40000 = 620 * 0.01

P = 248000 N or 2.48 *10^5 N or 248 kN

The problem involves calculating support reactions, shear force, axial force, and bending moment at mid-span of a beam under distributed and point loads using static equilibrium equations. It applies fundamentals of structural engineering, specifically focusing on shear force and bending moment analysis.

The question involves finding the support reactions at points A and B, and then determining the axial force (N), shear force (V), and bending moment (M) at the mid-span of the beam AB. Given the parameters are L = 4 m, q0 = 160 N/m, P = 200 N, and M0 = -380 N*m, the solution utilizes principles of static equilibrium to solve for the reactions caused by distributed and point loads on a statically determinate beam. This typically involves summing moments and forces in the horizontal and vertical directions.

Calculating these variables requires applying the fundamental equations of equilibrium, \(\Sigma F_x = 0\), \(\Sigma F_y = 0\), and \(\Sigma M = 0\), to the system. For example, the shear force (V) at any cross-section of the beam can be calculated by summing vertical forces, and the bending moment (M) can be calculated by summing moments about a point. The location of maximum shear force and bending moment often occurs at the supports or under the point load and varies linearly or parabolically along the beam length due to the distributed load.

The axial force (N) in this context is typically analyzed in members subjected to tension or compression, which might not be directly relevant for a simply supported beam under transverse loading unless it is specifically part of the problem's context, such as in the case of a truss member. In this problem, the main focus would be on shear force and bending moment calculations.

Answer:

The source code files for this question have been attached to this response.

Please download it and go through each of the class files.

The codes contain explanatory comments explaining important segments of the codes, kindly go through these comments.

Answer:

(a) Current density at P is [tex]J(P)=180.\textbf{a}_{\rho}-9.\textbf{a}_{\phi} \ (mA/m^2)\\[/tex].

(b) Total current I is 3.257 A

Explanation:

Because question includes symbols and formulas it can be misunderstood. In the question current density is given as below;

[tex]J=10\rho^2z.\textbf{a}_{\rho}-4\rho(\cos\phi)^2\textbf{a}_{\phi}\\[/tex]

where [tex]\textbf{a}_{\rho}[/tex] and [tex]\textbf{a}_{\phi}[/tex] unit vectors.

(a) In order to find the current density at a specific point (P), we can simply replace the coordinates in the current density equation.  Therefore

[tex]J(P(\rho=3, \phi=30^o,z=2))=10.3^2.2.\textbf{a}_{\rho}-4.3.(\cos(30^o)^2).\textbf{a}_{\phi}\\\\J(P)=180.\textbf{a}_{\rho}-9.\textbf{a}_{\phi} \ (mA/m^2)\\[/tex]

(b) Total current flowing outward can be calculated by using the relation,

[tex]I=\int {\textbf{J} \, \textbf{ds}[/tex]

where integral is calculated through the circular band given in the question. We can write the integral as below,

[tex]I=\int\{(10\rho^2z.\textbf{a}_{\rho}-4\rho(\cos\phi)^2\textbf{a}_{\phi}).(\rho.d\phi.dz.\textbf{a}_{\rho}})\}\\\\I=\int\{(10\rho^2z).(\rho.d\phi.dz)\}\\\\\\[/tex]

due to unit vector multiplication. Then,

[tex]I=10\int\(\rho^3z.dz.d\phi[/tex]

where [tex]\rho=3,\ 0<\phi<2\pi, \ 2<z<2.8[/tex]. Therefore

[tex]I=10.3^3\int_2^{2.8}\(zdz.\int_0^{2\pi}d\phi\\I=270(\frac{2.8^2}{2}-\frac{2^2}{2} )(2\pi-0)=3257.2\ mA\\I=3.257\ A[/tex]

Following are the solution to the given points:

Calculating the current density:

 [tex]\bold{J = 10 \rho^2 z a_{\rho} − 4 \rho \cos^2 \phi a_{\phi} \frac{mA}{m^2}}{}[/tex]

For point a:

Calculating the current density at point P:  

[tex]\to p=2 \\\\ \to \phi =60^{\circ}\\\\ \to Z=3[/tex]

[tex]\to J= 10(2)^2 3 (3) a_{\rho} - (4\times 2 \times ( \cos 60^{\circ})^2) a_{\phi}\\\\[/tex]

       [tex]=120 \ a_{\rho} - 4 \times 2 \times \frac{1}{2^2} a_{\phi} \\\\= 120 a_{\phi} - 2 a_{\phi} \frac{mA}{m^2}\\\\[/tex]

For point b:

Calculating the total current:  

[tex]\to I=\int \int \vec{J} \cdot \vec{ds}\\\\[/tex]

       [tex]= \int \int (10 \rho^2 z a_{\rho} − 4 \rho \cos^2 \phi a_{\phi}) \cdot (\rho d \phi \cdot dz) a_{\rho}[/tex]

Note:  

[tex]\text{p=constont then} \vec{ds} = \rho d \phi dz \hat{a}_{\rho} \\\\[/tex]

[tex]I= \int \int (10 \rho^2 z) \cdot ( d \phi dz)[/tex]

  [tex]= (10 \rho^3 \int^{8}_{3} z dz \int^{2 \pi}_{\pi} 1 \cdot d \phi\\\\= (10 \times 4^3 [\frac{z^2}{2}]^{8}_{3} \times [\phi]^{2 \pi}_{\pi}\\\\= 10 \times 64 \times \frac{(64-9)}{2} \times (2\pi-\pi) MA \\\\= 55292 \ MA \\\\= 55.292\ A[/tex]

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Answer:

The relative junction areas (A1 / A2) is 0.20

Explanation:

Using the formula of the collector current

ic = Is * [tex]e^{\frac{v_{be} }{V_{T} } }[/tex]

Is = [tex]\frac{ic}{e^{\frac{v_{be} }{V_{T} } }}[/tex]

Knowing

VT = 25 mV

vbe = 0.75 V

1 transistor

ic = 0.4 mA

Is1 = [tex]\frac{ic}{{\frac{v_{be} }{V_{T} } }}[/tex]

Is1 = [tex]\frac{0.4 * 10^{-3} }{{0.75}/{e^{25 * 10^{-3} } } }}[/tex]

Is1 = 5.47 * [tex]10^{-4}[/tex] A

2 transistor

ic = 2 mA

Is1 = [tex]\frac{ic}{{\frac{v_{be} }{V_{T} } }}[/tex]

Is1 = [tex]\frac{2 * 10^{-3} }{{0.75}/{e^{25 * 10^{-3} } } }}[/tex]

Is1 = 2.73 * [tex]10^{-3}[/tex] A

Junction area is proportional to saturation current

Is1 / Is2 = A1 / A2

A1 / A2 = 0.20