An isentropic steam turbine processes 5.5 kg/s of steam at 3 MPa, which is exhausted at 50 kPa and 100°C. Five percent of this flow is diverted for feedwater heating at 500 kPa. Determine the power produced by this turbine. Use steam tables.

Answer:

Total contraction on the bar = 1.238 mm

Explanation:

Modulus of Elasticity, E = 85 GN/m²

Diameter of the aluminium bar, [tex]d_{Al} = 40 mm = 0.04 m[/tex]

Load, P = 160 kN

Cross sectional area of the aluminium bar without hole:

[tex]A_1 = \frac{\pi d_{Al}^2 }{4} \\A_1 = \frac{\pi 0.04^2 }{4}\\A_1 = 0.00126 m^2[/tex]

Diameter of hole, [tex]d_h = 30 mm = 0.03 m[/tex]

Cross sectional area of the aluminium bar with hole:

[tex]A_2 = \frac{\pi( d_{Al}^2 - d_{h}^2)}{4} \\A_2 = \frac{\pi (0.04^2 - 0.03^2) }{4}\\A_2 = 0.00055 m^2[/tex]

Length of the aluminium bar, [tex]L_{Al} = 600 mm = 0.6 m[/tex]

Length of the hole, [tex]L_h = 100mm = 0.1 m[/tex]

Contraction in aluminium bar without hole  [tex]= \frac{P * L_{Al}}{A_1 E}[/tex]

Contraction in aluminium bar without hole  [tex]= \frac{160*10^3 * 0.6}{0.00126 * 85 * 10^9 }[/tex]

Contraction in aluminium bar without hole = 96000/107100000

Contraction in aluminium bar without hole = 0.000896

Contraction in aluminium bar with hole  [tex]= \frac{P * L_{h}}{A_2 E}[/tex]

Contraction in aluminium bar without hole  [tex]= \frac{160*10^3 * 0.1}{0.00055 * 85 * 10^9 }[/tex]

Contraction in aluminium bar without hole = 16000/46750000

Contraction in aluminium bar without hole = 0.000342

Total contraction = 0.000896 + 0.000342

Total contraction = 0.001238 m = 1.238 mm

Question:

A piston–cylinder device contains 0.85 kg of refrigerant- 134a at -10°C. The piston that is free to move has a mass of 12 kg and a diameter of 25 cm. The local atmospheric pressure is 88 kPa. Now, heat is transferred to refrigerant-134a until the temperature is 15°C. Determine (a) the final pressure, (b) the change in the volume of the cylinder, and (c) the change in the enthalpy of the refrigerant-134a.

Answer:

a) 90.4 kPa

b) 0.0205 m³

c) 17.4 kJ/kg

Explanation:

Given:

Mass, m = 0.85 kg

a) The final pressure here is equal to the initial pressure. Let's use the formula:

[tex] P_2 = P_1 = P_a_t_m + \frac{mg}{\pi D^2 / 4}[/tex]

[tex] = 88*10^3 + \frac{12kg * 9.81}{\pi (0.25)^2 / 4} [/tex]

= 90398 Pa

≈ 90.4 KPa

Final pressure = 90.4 kPa

b) Change in volume of the cylinder:

To find the initial and final volume, let's use the values from the A-13 table for refrigerant-134a, at initial values of 90.4 kPa and -10°C and final values of 90.4 kPa and 15°C

v1 = 0.2302m³/kg

h1 = 247.76 kJ/kg

v2 = 0.2544 m³/kg

h2 = 268.2 kJ/kg

Change in volume is calculated as:

Δv = m(v2 - v1)

Δv = 0.85(0.2544 - 0.2302)

= 0.0205 m³

Change in volume = 0.0205 m³

c) Change in enthalpy

Let's use the formula:

Δh = m(h2 - h1)

= 0.85(268.2 - 247.76)

= 17.4 kJ/kg

Change in enthalpy = 17.4 kJ/kg

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Answer:

The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.

Explanation:

The complete statement of this question is "Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320 MPa, and a factor of safety of 4.2 is required with respect to fracture. Determine the minimum allowable bolt diameter required to support an applied load of P = 450 kN"

Each bolt is subjected to shear forces. In this case, safety factor is the ratio of the ultimate shear strength to maximum allowable shear stress. That is to say:

[tex]n = \frac{S_{uts}}{\tau_{max}}[/tex]

Where:

[tex]n[/tex] - Safety factor, dimensionless.

[tex]S_{uts}[/tex] - Ultimate shear strength, measured in pascals.

[tex]\tau_{max}[/tex] - Maximum allowable shear stress, measured in pascals.

The maximum allowable shear stress is consequently cleared and computed: ([tex]n = 4.2[/tex], [tex]S_{uts} = 320\times 10^{6}\,Pa[/tex])

[tex]\tau_{max} = \frac{S_{uts}}{n}[/tex]

[tex]\tau_{max} = \frac{320\times 10^{6}\,Pa}{4.2}[/tex]

[tex]\tau_{max} = 76.190\times 10^{6}\,Pa[/tex]

Since each bolt has a circular cross section area and assuming the shear stress is not distributed uniformly, shear stress is calculated by:

[tex]\tau_{max} = \frac{4}{3} \cdot \frac{V}{A}[/tex]

Where:

[tex]\tau_{max}[/tex] - Maximum allowable shear stress, measured in pascals.

[tex]V[/tex] - Shear force, measured in kilonewtons.

[tex]A[/tex] - Cross section area, measured in square meters.

As connection consist on five bolts, shear force is equal to a fifth of the applied load. That is:

[tex]V = \frac{P}{5}[/tex]

[tex]V = \frac{450\,kN}{5}[/tex]

[tex]V = 90\,kN[/tex]

The minimum allowable cross section area is cleared in the shearing stress equation:

[tex]A = \frac{4}{3}\cdot \frac{V}{\tau_{max}}[/tex]

If [tex]V = 90\,kN[/tex] and [tex]\tau_{max} = 76.190\times 10^{3}\,kPa[/tex], the minimum allowable cross section area is:

[tex]A = \frac{4}{3} \cdot \frac{90\,kN}{76.190\times 10^{3}\,kPa}[/tex]

[tex]A = 1.640\times 10^{-3}\,m^{2}[/tex]

The minimum allowable cross section area can be determined in terms of minimum allowable bolt diameter by means of this expression:

[tex]A = \frac{\pi}{4}\cdot D^{2}[/tex]

The diameter is now cleared and computed:

[tex]D = \sqrt{\frac{4}{\pi}\cdot A}[/tex]

[tex]D =\sqrt{\frac{4}{\pi}\cdot (1.640\times 10^{-3}\,m^{2})[/tex]

[tex]D = 0.0457\,m[/tex]

[tex]D = 45.7\,mm[/tex]

The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.

We have that the minimum allowable bolt diameter is mathematically given as

From the question we are told

Generally the equation for the stress   is mathematically given as

[tex]\mu= 320/4.2 \\\\\mu= 76.190 N/mm^2[/tex]

Therefore

Force = Stress * area

Force = P/2

F= 425,000 N / 2 = 212,500 N

Hence area of each bolt is given as

212,500 = 76.190*( 5* area of each bolt)

area of each bolt = 557.815

Since

area of each bolt=\pi*d^2/4

\pi*d^2/4 = 557.815

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Answer:

Pressure difference (ΔP) = 63,519.75 kpa

Explanation:

Given:

ρ = 1,000 kg/m³

Height of cylindrical container used (h) = 7m / 2 = 3.5m

Specific gravity (sg) = 0.85

Find:

Pressure difference (ΔP).

Computation:

⇒ Pressure difference (ΔP) = h g [ ρ(sg) + ρ]                ∵ [ g = 9.81]

⇒ Pressure difference (ΔP) = (3.5)(9.81) [ 1,000(0.85) + 1,000]

⇒ Pressure difference (ΔP) = 34.335 [8,50 + 1,000]

⇒ Pressure difference (ΔP) = 34.335 [1,850]

⇒ Pressure difference (ΔP) = 63,519.75 kpa

Answer:

Current, I = 4A

Explanation:

Since the connection is in delta, let's convert to star.

Simplify BCD:

[tex] R1 = \frac{40 * 8}{40 + 16 + 8} = \frac{320}{64} = 5 ohms [/tex]

[tex] R2 = \frac{16 * 8}{40 + 16 + 8} = \frac{128}{64} = 2 ohms [/tex]

[tex] R3 = \frac{40 * 16}{40 + 16 + 8} = \frac{640}{64} = 10 ohms [/tex]

From figure B, it can be seen that 6 ohms and 6 ohms are connected in parallel.

Simplify:

[tex] \frac{6 * 6}{6 + 6} = \frac{36}{12} = 3 \ohms [/tex]

Req = 10 ohms + 3 ohms

Req = 13 ohms

To find the current, use ohms law.

V = IR

Where, V = 52volts and I = 13 ohms

Solve for I,

[tex] I = \frac{V}{R} = \frac{52}{13} = 4A[/tex]

Current, I = 4 A

Answer:

Explanation:

Consider schedules S3, S4, and S5 below. Determine whether each schedule is strict, cascadeless, recoverable, or non-recoverable. You need to explain your reason.

S3: r1(x), r2(z), r1(z), r3(x), r3(y), w1(x), c1, w3(y), c3, r2(y), w2(z),w2(y),c2

S4: r1(x), r2(z), r1(z), r3(x), r3(y),w1(x),w3(y), r2(y),w2(z),w2(y), c1,c2, c3

S5: r1(x), r2(z), r3(x), r1(z), r2(y), r3(y), w1(x), c1, w2(z), w3(y), w2(y), c3, c2

Strict schedule:

A schedule is strict if it satisfies the following conditions:

Tj reads a data item X after Ti has written to X and Ti is terminated means aborted or committed.

Tj writes a data item X after Ti has written to X and Ti is terminated means aborted or committed.

S3 is not strict because In a strict schedule T3 must read X after C1 but here T3 reads X (r3(X)) before Then T1 has written to X (w1(X)) and T3 commits after T1.

S4 is not strict because In a strict schedule T3 must read X after C1, but here T3 reads X (r3(X)) before T1 has written to X (w1(X)) and T3 commits after T1.

S5 is not strict because T3 reads X (r3(X)) before T1 has written to X (w1(X))

but T3 commits after T1. In a strict schedule T3 must read X after C1.

Cascadeless schedule:

Cascadeless schedule follows the below condition:

Tj reads X only? after Ti has written to X and terminated means aborted or committed.

S3 is not cascadeless schedule because T3 reads X (r3(X)) before T1 commits.

S4 is not cascadeless schedule because T3 reads X (r3(X)) before T1 commits.

S5 is not cascadeless schedule because T3 reads X (r3(X)) before T1 commits or T2 reads Y (r2(Y)) before T3 commits.

But while come to the definition of cascadeless schedules S3, S4, and S4 are not cascadeless, and T3 is not affected if T1 is rolled back in any of the schedules, that is,

T3 does not have to roll back if T1 is rolled back. The problem occurs because these

schedules are not serializable.

Recoverable schedule:

Schedule that follows the below condition:

-----Tj commits after Ti if Tj has?read any data item written by Ti.

Ci > Cj means that Ci happens before Cj. Ai denotes abort Ti. To test if a schedule is

recoverable one has to include abort operations. Thus in testing the recoverability abort

operations will have to used in place of commit one at a time. Also the strictest condition is

------where a transaction neither reads nor writes to a data item, which was written to by a transaction that has not committed yet.

If A1?>C3>C2, then schedule S3 is recoverable because rolling back of T1 does not affect T2 and

T3. If C1>A3>C2. schedule S3 is not recoverable because T2 read the value of Y (r2(Y)) after T3 wrote X (w3(Y)) and T2 committed but T3 rolled back. Thus, T2 used non- existent value of Y. If C1>C3>A3, then S3 is recoverable because roll back of T2 does not affect T1 and T3.

Strictest condition of schedule S3 is C3>C2.

If A1?>C2>C3, then schedule S4 is recoverable because roll back of T1 does not affect T2 and T3. If C1>A2>C3, then schedule S4 is recoverable because the roll back of T2 will restore the value of Y that was read and written to by T3 (w3(Y)). It will not affect T1. If C1>C2>A3, then schedule S4 is not recoverable because T3 will restore the value of Y which was not read by T2.

Answer:

A.) Time = 17.13 seconds

B.) Distance = 31.9 m

C.) V = 11.18 m/s

D.) V = 7.1 m/s

Explanation:

The initial velocity U of the automobile is 15.65 m/s.

 At the instant that the two vehicles are abreast of each other, the police car starts to pursue the automobile with initial velocity U = 0 at a constant acceleration of 1.96 m/s². Because the police is starting from rest.

For the automobile, let us use first equation of motion

V = U - at.

Acceleration a is negative since it is decelerating with a = 3.05 m/s² . And

V = 0.

Substitute U and a into the formula

0 = 15.65 - 3.05t

15.65 = 3.05t

t = 15.65/3.05

t = 5.13 seconds

But the motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s².

The total time required for the police car to overtake the automobile will be

12 + 5.13 = 17.13 seconds.

b.) Using the third equation of motion formula for the police car at V = 11.18 m/s and a = 1.96 m/s²

V^2 = U^2 + 2aS

Where S = distance travelled.

Substitute V and a into the formula

11.18^2 = 0 + 2 × 1.96 ×S

124.99 = 3.92S

S = 124.99/3.92

S = 31.88 m

c.) The speed of the police car at the time it overtakes the automobile will be in line with the speed zone which is 11.18 m/s

d.) That will be the final velocity V of the automobile car.

We will use third equation of motion to solve that.

V^2 = U^2 + 2as

V^2 = 15.65^2 - 2 × 3.05 × 31.88

V^2 = 244.9225 - 194.468

V = sqrt( 50.4545)

V = 7.1 m/s

Answer:

Develop social skills

Explanation:

Answer:

strengthen your college applications

Explanation:

Answer:

With the Breakthrough of Technology, the rate at which things are done are becoming much more easy. but without basic science, innovation towards technology cannot occur, so the both work hand in hand in the world of technology today.

Explanation:

Technological innovation and Basic science research plays a major role in the world of science and technology today, while we all want technology innovation the more, without basic science, innovation cannot come in place,

Just as we are going further in technology, breakthroughs and growth are been made which helps on the long run in science research which in turn has made things to be done much better and easily.

Answer:

(i) IR = 100.167 A Iy = 75.125∠-120 IB = 50.083 ∠+120 (ii) IN =43.374∠ -30°

Explanation:

Solution

Given that:

Three loads  24 kW, 18 kW, and 12 kW are connected between the neutral.

Voltage = 415V

Now,

(1)The current in each line conductor

Thus,

The Voltage Vpn = vL√3

Gives us, 415/√3 = 239.6 V

Then,

IR = 24 K/ Vpn ∠0°

24K/239.6 ∠0°= 100.167 A

For Iy

Iy = 18k/239. 6

= 75.125A

Thus,

Iy = 75.125∠-120 this is as a result of the 3- 0 system

Now,

IB = 12K /239.6

= 50.083 A

Thus,

IB is =50.083 ∠+120

(ii) We find the current in the neutral conductor

which is,

IN =Iy +IB +IR

= 75.125∠-120 + 50.083∠+120 +100.167

This will give us the following summation below:

-37.563 - j65.06 - 25.0415 +j 43.373 + 100.167

Thus,

IN = 37.563- j 21.687

Therefore,

IN =43.374∠ -30°

Answer:

Both technician A and technician B are correct

Explanation:

A planetary gearbox consists of a gearbox with the input shaft and the output shaft that is aligned to each other. It is used to transfer the largest torque in the compact form. A planetary gearbox has a compact size and low weight and it has high power density.

One planetary gear set can provide gear reduction, overdrive, and reverse. Also, most transmissions today use compound (multiple) planetary gears set.

So, both technician A and technician B are correct.

Answer:

Explanation:

Given that:

V = 12.5m/s

L= 2.70m

b= 0.65m

[tex]T_{ \infty} = 27^0C= 273+27 = 300K[/tex]

[tex]T_s= 127^0C = (127+273)= 400K[/tex]

P = 1atm

Film temperature

[tex]T_f = \frac{T_s + T_{\infty}}{2} \\\\=\frac{400+300}{2} \\\\=350K[/tex]

dynamic viscosity =

[tex]\mu =20.9096\times 10^{-6} m^2/sec[/tex]

density = 0.9946kg/m³

Pr = 0.708564

K= 229.7984 * 10⁻³w/mk

Reynolds number,

[tex]Re = \frac{SUD}{\mu} =\frac{\ SUl}{\mu}[/tex]

[tex]=\frac{0.9946 \times 12.5\times 2.7}{20.9096\times 10^-^6} \\\\Re=1605375.043[/tex]

we have,

[tex]Nu=\frac{hL}{k} =0.037Re^{4/5}Pr^{1/3}\\\\\frac{h\times2.7}{29.79\times 10^-63} =0.037(1605375.043)^{4/5}(0.7085)^{1/3}\\\\h=33.53w/m^2k[/tex]

we have,

heat transfer rate from top plate

[tex]\theta _1 =hA(T_s-T_{\infty})\\\\A=Lb\\\\=2.7*0.655\\\\ \theta_1=33.53*2.7*0.65(127/27)\\\\ \theta_1=5884.51w[/tex]

Answer:

Settling Velocity (Up)= 2.048*10^-5 m/s

Reynolds number Re = 2.159*10^-3

Explanation:

We proceed as follows;

Diameter of Particle = 0.09 mm = 0.09*10^-3 m

Solid Particle Density = 2002 kg/m3

Solid Fraction, θ= 0.45

Temperature = 26.7°C

Viscosity of water = 0.8509*10^-3 kg/ms

Density of water at 26.7 °C = 996.67 kg/m3

The velocity between the interface, i.e between the suspension and clear water is given by,

U = [ ((nf/ρf)/d)D^3] [18+(1/3)D^3)(1/2)]

D = d[(ρp/ρf)-1)g*(ρf/nf)^2]^(1/3)

D = 2.147

U = 0.0003m/s (n = 4.49)

Up = 0.0003 * (1-0.45)^4.49 = 2.048*10^-5 m/s

Re=0.09*10^-3*2.048*10^-5*996.67/0.0008509 = 2.159*10^-3

Answer:

1) 2304 kPa

2) B. 200 N/m

Explanation:

The internal pressure of the of the tank  can be found from the following relations;

Resisting wall force F = p×(1/4·π·D²)

σ×A = p×(1/4·π·D²)

Where:

σ = Allowable stress of the tank

A = Area of the wall of the tank = π·D·t

t = Thickness of the tank = 15 mm. = 0.015 m

D = Diameter of the tank = 25 m

p = Maximum permissible internal pressure pressure

∴ σ×π·D·t = p×(1/4·π·D²)

p = 4×σ×t/D = 4 × 240 ×0.015/2.5 = 5.76 MPa

With a desired safety factor of 2.5, the permissible internal pressure = 5.76/2.5 = 2.304 MPa

2) The formula for average shear flow is given as follows;

[tex]q = \dfrac{T}{2 \times A_m}[/tex]

Where:

q = Average shear flow

T = Torque = 8 N·m

[tex]A_m[/tex] = Average area enclosed within tube

t = Thickness of tube = 6.35 mm = 0.00635 m

Side length of the square cross sectioned tube, s = 203 mm = 0.203 m

Average area enclosed within tube, [tex]A_m[/tex] = (s - t)² = (0.203 - 0.00635)² = 0.039 m²

[tex]\therefore q = \dfrac{8}{2 \times 0.039} = 206.9 \, N/m[/tex]

Hence the average shear flow is most nearly 200 N/m.

Following are the solution to the given question:

Calculating the allowable stress:

[tex]\to \sigma_{allow} = \frac{\sigma_y}{FS} \\\\[/tex]

              [tex]= \frac{240}{2.5} \\\\= 96\\\\[/tex]

Calculating the Thickness:

[tex]\to t =15\ mm = \frac{15\ }{1000}= 0.015\ m\\\\[/tex]

The stress in a spherical tank is defined as

[tex]\to \sigma = \frac{pD}{4t}\\\\\to 96 = \frac{p(25)}{4(0.015)}\\\\\to p = 0.2304\;\;MPa\\\\\to p = 230.4\;\;kPa\\\\\to p \approx 230\;\;kPa\\\\[/tex]

[tex]\bold{\to A= 203^2= 41209\ mm^2} \\\\[/tex]

Calculating the shear flow:

[tex]\to q=\frac{T}{2A}[/tex]

      [tex]=\frac{8}{2 \times 41209 \times 10^{-6}}\\\\=\frac{8}{0.082418}\\\\=97.066\\\\[/tex]

[tex]\to q=97 \approx 100 \ \frac{N}{m}\\[/tex]

Therefore, the final answer is "".

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Answer:

Yes, the forging can be completed

Explanation:

Given h = 100 mm, ε = ㏑(450/100) = 1.504

[tex]Y_f = 230 \times 1.504^{0.15} = 244.52[/tex]

V = π·D²·L/4 = π × 50²×450/4 = 883,572.93 mm³

At h = 100 mm, A = V/h = 883,572.93 /100 = 8835.73 mm²

D = √(4·A/π) = 106.07 mm

[tex]K_f[/tex] = 1 + 0.4 × 0.1 × 106.07/100 = 1.042

F = 1.042 × 244.52 × 8835.73 = 2252199.386 N =2.25 MN

Hence the required force = 2.25 MN is less than the available force = 3 MN therefore, the forging can be completed.

Answer:

See Explanation

Explanation:

Required

- Pseudocode to determine the largest of 10 numbers

- C# program to determine the largest of 10 numbers

The pseudocode and program makes use of a 1 dimensional array to accept input for the 10 numbers;

The largest of the 10 numbers is then saved in variable Largest and printed afterwards.

Pseudocode (Number lines are used for indentation to illustrate the program flow)

1. Start:

2. Declare Number as 1 dimensional array of 10 integers

3. Initialize: counter = 0

4. Do:

4.1 Display “Enter Number ”+(counter + 1)

4.2 Accept input for Number[counter]

4.3 While counter < 10

5. Initialize: Largest = Number[0]

6. Loop: i = 0 to 10

6.1 if Largest < Number[i] Then

6.2 Largest = Number[i]

6.3 End Loop:

7. Display “The largest input is “+Largest

8. Stop

C# Program (Console)

Comments are used for explanatory purpose

using System;

namespace ConsoleApplication1

{

   class Program

   {

       static void Main(string[] args)

       {

           int[] Number = new int[10];  // Declare array of 10 elements

           //Accept Input

           int counter = 0;

           while(counter<10)

           {

               Console.WriteLine("Enter Number " + (counter + 1)+": ");

               string var = Console.ReadLine();

               Number[counter] = Convert.ToInt32(var);

               counter++;                  

           }

           //Initialize largest to first element of the array

           int Largest = Number[0];

           //Determine Largest

           for(int i=0;i<10;i++)

           {

               if(Largest < Number[i])

               {

                   Largest = Number[i];

               }

           }

           //Print Largest

           Console.WriteLine("The largest input is "+ Largest);

           Console.ReadLine();

       }

   }

}

Answer:

a) [tex]\ddot n = 50400\,\frac{rev}{min^{2}}[/tex], b) [tex]n = 2450\,rev[/tex]

Explanation:

a) The acceleration experimented by the grinding wheel is:

[tex]\ddot n = \frac{4200\,\frac{rev}{min} - 0 \,\frac{rev}{min} }{\frac{5}{60}\,min }[/tex]

[tex]\ddot n = 50400\,\frac{rev}{min^{2}}[/tex]

Now, the number of revolutions done by the grinding wheel in that period of time is:

[tex]n = \frac{(4200\,\frac{rev}{min} )^{2}-(0\,\frac{rev}{min} )^{2}}{2\cdot \left(50400\,\frac{rev}{min^{2}} \right)}[/tex]

[tex]n = 175\,rev[/tex]

b) The acceleration experimented by the grinding wheel is:

[tex]\ddot n = \frac{0 \,\frac{rev}{min} - 4200\,\frac{rev}{min} }{\frac{70}{60}\,min }[/tex]

[tex]\ddot n = -3600\,\frac{rev}{min^{2}}[/tex]

Now, the number of revolutions done by the grinding wheel in that period of time is:

[tex]n = \frac{(0\,\frac{rev}{min} )^{2} - (4200\,\frac{rev}{min} )^{2}}{2\cdot \left(-3600\,\frac{rev}{min^{2}} \right)}[/tex]

[tex]n = 2450\,rev[/tex]

Answer:

s = v√[2(H - h)/g]

This formula describes the distance from the building that Prof Murthy must be when you release the egg

Explanation:

First, we need to find the time required by the egg to reach the head of Professor. For that purpose, we use 1st equation of motion in vertical form:

Vf = Vi + gt

where,

Vf = Velocity of egg at the time of hitting head of the Professor

Vi = initial velocity of egg = 0 m/s  (Since, egg is initially at rest)

g = acceleration due to gravity

t = time taken by egg to come down

Therefore,

Vf = 0 + gt

t = Vf/g   ----- equation (1)

Now, we use 3rd euation of motion for Vf:

2gs = Vf² - Vi²

where,

s = height dropped = H - h

Therefore,

2g(H - h) = Vf²

Vf = √[2g(H - h)]

Therefore, equation (1) becomes:

t = √[2g(H - h)]/g

t = √[2(H - h)/g]

Now, consider the horizontal motion of professor. So, the minimum distance of professor from building can be found out by finding the distance covered by the professor in time t. Since, the professor is running at constant speed of v m/s. Therefore:

s = vt

s = v√[2(H - h)/g]

This formula describes the distance from the building that Prof Murthy must be when you release the egg

Answer:

The dimension of the  square cross section is = 30mm * 30mm

Explanation:

Before proceeding with the calculations convert MPa to Kpsi

Sut ( ultimate strength ) = 770 MPa * 0.145 Kpsi/MPa = 111.65 Kpsi

the fatigue strength factor of Sut at 10^3 cycles for Se = Se' = 0.5 Sut

at 10^6 cycles" for 111.65 Kpsi  = f ( fatigue strength factor) = 0.83

To calculate the endurance limit  use  Se' = 0.5 Sut      since Sut < 1400 MPa

Se'( endurance limit ) = 0.5 * 770 = 385 Mpa

The surface condition modification factor

Ka = 57.7 ( Sut )^-0.718

Ka = 57.7 ( 770 ) ^-0.718

Ka = 0.488

Assuming the size modification factor (Kb) = 0.85 and also assuming all modifiers are equal to one

The endurance limit at the critical location of a machine part can be expressed as :

Se = Ka*Kb*Se'

Se = 0.488 * 0.85 * 385 = 160 MPa

Next:

Calculating the constants to find the number of cycles

α = [tex]\frac{(fSut)^2}{Se}[/tex]

α =[tex]\frac{(0.83*770)^2}{160}[/tex]  =  2553 MPa

b = [tex]-\frac{1}{3} log(\frac{fSut}{Se} )[/tex]

b = [tex]-\frac{1}{3} log (\frac{0.83*770}{160} )[/tex]  = -0.2005

Next :

calculating the fatigue strength using the relation

Sf = αN^b

N = number of cycles

Sf = 2553 ( 10^4) ^ -0.2005

Sf = 403 MPa

Calculate the maximum moment of the beam

M = 2000 * 0.6 = 1200 N-m

calculating the maximum stress in the beam

∝ = ∝max = [tex]\frac{Mc}{I}[/tex]

Where c = b/2 and   I = b(b^3) / 12

hence ∝max = [tex]\frac{6M}{b^3}[/tex]  =  6(1200) / b^3   =  7200/ b^3   Pa

note: b is in (meters)

The expression for the factor of safety is written as

n = [tex]\frac{Sf}{\alpha max }[/tex]

Sf = 403, n = 1.5 and ∝max = 7200 / b^3

= 1.5 = [tex]\frac{403(10^6 Pa/Mpa)}{7200 / B^3}[/tex]   making b subject of the formula in other to get the value of b

b = 0.0299 m * 10^3 mm/m

b = 29.9 mm therefore b ≈ 30 mm

since  the size factor  assumed is near the calculated size factor using this relation : de = 0.808 ( hb)^1/2

the dimension = 30 mm by 30 mm

Answer:

A. True

Explanation:

When an electromagnetic field wave strikes a conductor, say a wire, it induces an alternating current that is proportional to the wave in the conductor. This is a reversal of generating electromagnetic wave from accelerating a charged particle. This phenomenon is used in radio antena for receiving radio wave signals and also use in medicine for body scanning.

Answer:

final temperature = 26.5°

Explanation:

Initial volume of water is 1 x 1 x 1 = 1 [tex]m^{3}[/tex]

Initial temperature of water = 20° C

Density of water = 1000 kg/[tex]m^{3}[/tex]

volume of copper block = 0.46 x 0.46 x 0.46 = 0.097 [tex]m^{3}[/tex]

Initial temperature of copper block = 100° C

Density of copper = 8960 kg/[tex]m^{3}[/tex]

Final volume of water = 1 - 0.097 = 0.903 [tex]m^{3}[/tex]

Assumptions:

mass of water remaining in the tank will be density x volume = 1000 x 0.903 = 903 kg

specific heat capacity of water c = 4186 J/K-kg

heat content of water left Hw = mcT = 903 x 4186 x 20 = 75.59 Mega-joules

mass of copper will be density x volume = 8960 x 0.097 = 869.12 kg

specific heat capacity of copper is 385 J/K-kg

heat content of copper Hc = mcT = 869.12 x 385 x 100 = 33.46 Mega-joules

total heat in the system = 75.59 + 33.46 = 109.05 Mega-joules

this heat will be distributed in the entire system

heat energy of water within the system = mcT

where T is the final temperature

= 903 x 4186 x T = 3779958T

for copper, heat will be

mcT = 869.12 x 385 = 334611.2T

these component heats will sum up to the final heat of the system, i.e

3779958T + 334611.2T = 109.05 x [tex]10^{6}[/tex]

4114569.2T = 109.05 x [tex]10^{6}[/tex]

final temperature T = (109.05 x [tex]10^{6}[/tex])/4114569.2 = 26.5°

Answer:

The minimum diameter to withstand such tensile strength is 22.568 mm.

Explanation:

The allow is experimenting an axial load, so that stress formula for cylidrical sample is:

[tex]\sigma = \frac{P}{A_{c}}[/tex]

[tex]\sigma = \frac{4\cdot P}{\pi \cdot D^{2}}[/tex]

Where:

[tex]\sigma[/tex] - Normal stress, measured in kilopascals.

[tex]P[/tex] - Axial load, measured in kilonewtons.

[tex]A_{c}[/tex] - Cross section area, measured in square meters.

[tex]D[/tex] - Diameter, measured in meters.

Given that [tex]\sigma = 75\times 10^{3}\,kPa[/tex] and [tex]P = 30\,kN[/tex], diameter is now cleared and computed at last:

[tex]D^{2} = \frac{4\cdot P}{\pi \cdot \sigma}[/tex]

[tex]D = 2\sqrt{\frac{P}{\pi \cdot \sigma} }[/tex]

[tex]D = 2 \sqrt{\frac{30\,kN}{\pi \cdot (75\times 10^{3}\,kPa)} }[/tex]

[tex]D = 0.0225\,m[/tex]

[tex]D = 22.568\,mm[/tex]

The minimum diameter to withstand such tensile strength is 22.568 mm.

Image of wheel is missing, so i attached it.

Answer:

ω = 14.95 rad/s

Explanation:

We are given;

Mass of wheel; m = 20kg

T = 20 N

k_o = 0.3 m

Since the wheel starts from rest, T1 = 0.

The mass moment of inertia of the wheel about point O is;

I_o = m(k_o)²

I_o = 20 * (0.3)²

I_o = 1.8 kg.m²

So, T2 = ½•I_o•ω²

T2 = ½ × 1.8 × ω²

T2 = 0.9ω²

Looking at the image of the wheel, it's clear that only T does the work.

Thus, distance is;

s_t = θr

Since 4 revolutions,

s_t = 4(2π) × 0.4

s_t = 3.2π

So, Energy expended = Force x Distance

Wt = T x s_t = 20 × 3.2π = 64π J

Using principle of work-energy, we have;

T1 + W = T2

Plugging in the relevant values, we have;

0 + 64π = 0.9ω²

0.9ω² = 64π

ω² = 64π/0.9

ω = √64π/0.9

ω = 14.95 rad/s

Answer:

Check the v(t) signal referred to in the question and the solution to each part in the files attached

Explanation:

The detailed solutions of parts a to d are clearly expressed in the second file attached.

Answer:

A periodic function is a function that returns to its value over a certain period at regular intervals an example is the wave form of flux density (B) = sin wt

Explanation:

A periodic function is a function that returns to its value over a certain period at regular intervals an example is the wave form of flux density (B) = sin wt

attached to the answer is a free plot of the output starting with zero degree for one coil rotating in a uniform magnetic field

B ( wave flux density ) = Bm sinwt  and w = 2[tex]\pi[/tex]f = [tex]\frac{2\pi }{T}[/tex] rad/sec

Answer:

1) The normal reactions at the front wheel is 9909.375 N

The normal reactions at the rear wheel is 8090.625 N

2) The least coefficient of friction required between the tyres and the road is 0.625

Explanation:

1) The parameters given are as follows;

Speed, u = 90 km/h = 25 m/s

Distance, s it takes to come to rest = 50 m

Mass, m = 1.8 tonnes = 1,800 kg

From the equation of motion, we have;

v² - u² = 2·a·s

Where:

v = Final velocity = 0 m/s

a = acceleration

∴ 0² - 25² = 2 × a × 50

a = -6.25 m/s²

Force, F =  mass, m × a = 1,800 × (-6.25) = -11,250 N

The coefficient of friction, μ, is given as follows;

[tex]\mu =\dfrac{u^2}{2 \times g \times s} = \dfrac{25^2}{2 \times 10 \times 50} = 0.625[/tex]

Weight transfer is given as follows;

[tex]W_{t}=\dfrac{0.625 \times 0.9}{3}\times \dfrac{6.25}{10}\times 18000 = 2109.375 \, N[/tex]

Therefore, we have for the car at rest;

Taking moment about the Center of Gravity CG;

[tex]F_R[/tex] × 1.3 = 1.7 × [tex]F_F[/tex]

[tex]F_R[/tex] + [tex]F_F[/tex] = 18000

[tex]F_R + \dfrac{1.3 }{1.7} \times F_R = 18000[/tex]

[tex]F_R[/tex] = 18000*17/30 = 10200 N

[tex]F_F[/tex] = 18000 N - 10200 N = 7800 N

Hence with the weight transfer, we have;

The normal reactions at the rear wheel [tex]F_R[/tex]  = 10200 N - 2109.375 N = 8090.625 N

The normal reactions at the front wheel [tex]F_F[/tex] =  7800 N + 2109.375 N = 9909.375 N

2) The least coefficient of friction, μ, is given as follows;

[tex]\mu = \dfrac{F}{R} = \dfrac{11250}{18000} = 0.625[/tex]

The least coefficient of friction, μ = 0.625.

Answer:

[tex] T_A_C = 6.296 kN [/tex]

[tex] R = 10.06 kN [/tex]

Explanation:

Given:

[tex] T_A_B = 8.7 kN[/tex]

Required:

Find the tension TAC and magnitude R of this downward force.

First calculate [tex] \alpha, \beta, \gamma [/tex]

[tex] \alpha = tan^-^1 =\frac{40}{50} = 38. 36 [/tex]

[tex] \beta = tan^-^1 =\frac{50}{30} = 59.04 [/tex]

[tex] \gamma = 180 - 38.36 - 59.04 = 82.6 [/tex]

To Find tension in AC and magnitude R, use sine rule.

[tex] \frac{sin a}{T_A_C} = \frac{sin b}{T_A_B} = \frac{sin c}{R} [/tex]

Substitute values:

[tex]\frac{sin 38.36}{T_A_C} = \frac{sin 59.04}{8.7} = \frac{82.6}{R}[/tex]

Solve for T_A_C:

[tex] T_A_C = 8.7 * \frac{sin 38.36}{sin 59.04} = [/tex]

[tex] T_A_C = 8.7 * 0.724 = 6.296 kN [/tex]

Solve for R.

[tex] R = 8.7 * \frac{sin 82.6}{sin 59.04} = [/tex]

[tex] R = 8.7 * 1.156 [/tex]

R = 10.06 kN

Tension AC = 6.296kN

Magnitude,R = 10.06 kN