A float valve, regulating the flow of water into a reservoir, is shown in the figure. The spherical float (half of the sphere is submerged) is 0.1553 m in diameter. AOB is the weightless link carrying the float at one end, and a valve at the other end which closes the pipe through which flows into the reservoir. The link is mounted on a frictionless hinge at O, and the angle AOB is 135o. The length of OA is 253 mm and the distance between the center of the float and the hinge is 553 mm. When the flow is stopped AO will be vertical. The valve is to be pressed on to the seat with a force of 10,53 N to be completely stop the flow in the reservoir. It was observed that the flow of water is stopped, when the free surface of water in the reservoir is 353 mm below the hinge at O. Determine the weight of the float sphere.

Answer:

HB = 3.22

Explanation:

The formula to calculate the Brinell Hardness is given as follows:

[tex]HB = \frac{2P}{\pi D\sqrt{D^{2}- d^{2} } }[/tex]

where,

HB = Brinell Hardness = ?

P = Applied Load in kg = 500 kg

D = Diameter of Indenter in mm = 10 mm

d = Diameter of the indentation in mm = 1.55 mm

Therefore, using these values, we get:

[tex]HB = \frac{(2)(500)}{\pi (10)\sqrt{10^{2}- 1.55^{2} } }[/tex]

HB = 3.22

Answer:

Explanation:

For cylinder

Diameter d = 60 inches

thickness t = 0.625 inches

circumferential (hoop) stress = 30 ksi

[tex]hoop \ \ stress =\sigma_1=\frac{P_1d}{2t}\\\\\sigma_1=30ksi\\\\30000=\frac{P_1\times 60}{2\times0.625}\\\\P_1=624psi[/tex]

[tex]longitudinal \ \ stress =\sigma_2=\frac{P_2d}{2t}\\\\\sigma_2=30ksi\\\\30000=\frac{P_2\times 60}{4\times0.625}\\\\30000=\frac{P_2\times 60}{2.5}\\\\75000=P_2\times60\\\\P_2=\frac{75000}{60} \\\\P_1=1250psi[/tex]

Therefore maximum pressure without failure is P₁ = 625 psi

ii) For Sphere

[tex]\sigma_1=\sigma_2=\frac{Pd}{4t} \\\\P=\frac{30000\times 4 \times 0.625}{60} \\\\=\frac{75000}{60}\\\\=1250\ \ psi[/tex]

Answer:

Technician B is correct

Explanation:

O- rings are used with oil transmission filters to avoid transmission failures but some people use  lip seals as well. either of them is  inserted onto the outer part of the transmission system i.e it is inserted/found in-between Transmission filters and the transmission systems and it main purpose is to avoid leaks and transmission failure in the short and long term.

0-rings should be lubricated before installation this is because the o-rings are usually super tight when installing and would require lubrication to ease the installation process else the rubber might get ruptured and this would lead to instant transmission failure.

Answer:

(a) 12 seconds (b) t = 8.31 seconds (c) 10µ A (d) V = 20 V (e) V =0

Explanation:

Solution

Given that:

C = 6 µ which is = 6 * 10^ ⁻6

R = 2 MΩ, which is = 2 * 10^ 6

ε = 20 V

(a) When it is at the time constant we have the following:

λ = CR

= 6 * 10^ ⁻6 * 2 * 10^ 6

λ =12 seconds

(b) We solve for the half life of the circuit which is given below:

d₀ = d₀ [ 1- e ^ ⁺t/CR

d = decay mode]

d₀/2 =  d₀  1- e ^ ⁺t/12

2^⁻1 = e ^ ⁺t/12

Thus

t/12 ln 2

t = 12 * ln 2

t = 12 * 0.693

t = 8.31 seconds

(c) We find the current at t = 0

So,

I = d₀/dt

I = d₀/dt e ^ ⁺t/CR

= CE/CR e ^ ⁺t/CR

E/R e ^ ⁺t/CR

Thus,

at t = 0

I  E/R = 20/  2 * 10^ 6

= 10µ A

(d) We find the voltage across the capacitor at t = 0 which is shown below:

V = IR

= 10 * 10^ ⁻6 * 2 * 10^ 6

V = 20 V

(e)  We solve for he voltage across the resistor.

At t = 0

I = 0

V =0

Answer:

the % recovery of aluminum product is 80.5%

the % purity of the aluminum product is 54.7%

Explanation:

feed rate to separator = 2500 kg/hr

in one hour, there will be 2500 kg/hr x 1 hr = 2500 kg of material is fed into the  machine

of this 2500 kg, the feed is known to contain 174 kg of aluminium and 2326 kg of rejects.

After the separation, 256 kg  is collected in the product stream.

of this 256 kg, 140 kg is aluminium.

% recovery of aluminium will be = mass of aluminium in material collected in the product stream ÷ mass of aluminium contained in the feed material

% recovery of aluminium = 140kg/174kg x 100% = 80.5%

% purity of the aluminium product = mass of aluminium in final product ÷ total mass of product collected in product stream

% purity of the aluminium product = 140kg/256kg

x 100% = 54.7%

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:

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:

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:

Exit temperature = 32 °C

Explanation:

We are given;

Initial Pressure;P1 = 100 KPa

Cp =1000 J/kg.K = 1 KJ/kg.k

R = 500 J/kg.K = 0.5 Kj/Kg.k

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

volume flow rate;V' = 15 m³/s

W = 130 Kw

Q = 80 Kw

Using ideal gas equation,

PV' = m'RT

Where m' is mass flow rate.

Thus;making m' the subject, we have;

m' = PV'/RT

So at inlet,

m' = P1•V1'/(R•T1)

m' = (100 × 15)/(0.5 × 300)

m' = 10 kg/s

From steady flow energy equation, we know that;

m'•h1 + Q = m'h2 + W

Dividing through by m', we have;

h1 + Q/m' = h2 + W/m'

h = Cp•T

Thus,

Cp•T1 + Q/m' = Cp•T2 + W/m'

Plugging in the relevant values, we have;

(1*300) - (80/10) = (1*T2) - (130/10)

Q and M negative because heat is being lost.

300 - 8 + 13 = T2

T2 = 305 K = 305 - 273 °C = 32 °C

13000 + 300 - 8000 = T2

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.

Bank loan facilitator, and hospital emergency care specialist are the two consumer or customer services careers.

Learn more about customer:

https://brainly.com/question/13735743

The common" concern with using variable frequency drives (VFDs) is C. hertz modulation.

It should be noted that a variable frequency drive simply means a type of motor drive that us used in mechanical drive system.

In this case, common" concern with using variable frequency drives (VFDs) is hertz modulation

Learn more about frequency on:

brainly.com/question/6985885

#SPJ9

Answer:

(i) The inductance of the inductor is = 43.43 mH (ii) the impedance of the circuit is = 16∠58.61° Ω (iii) the phase difference for current and the voltage applied is Q = 58.61°

Explanation:

Solution

Given that:

I= 5 A

V = 125V

Resistance R= Not known yet

Thus

To find the resistance we have the following formula which is shown below:

R = V/I

=125/15

R =8.333Ω

Now,

Voltage = 240

Frequency = 50Hz

Current (I) remain at = 15A

Z= not known (impedance)

so,

To find the impedance we have the formula which is shown below:

Z = V/I =240/15

Z= 16Ω⇒ Z = R + jXL

Z = 8.333 + jXL = 16

Thus

√8.333² + XL² = 16²

8.333² + XL² = 16²

XL² = 186.561

XL = 13.658Ω

Now

We find the inductance of the Inductor and the impedance of the circuit.

(i) In solving for the inductance of the inductor, a formula is applied here, which is shown below:

L =  XL/w

=13.658/ 2π * 50

=13.658/314.15 = 0.043 = 43.43 mH

Note: w= 2πf

(ii) For the impedance of the circuit we have the following:

z = 8.333 + j 13.658

z = 16∠58.61° Ω

(iii) The next step is to find the phase difference between the applied voltage and current.

Q =  this is the voltage across the inductor in a series of resonant circuit.

Q can also be called the applied voltage

Thus,

Q is described as an Impedance angle

Therefore, Q = 58.81°

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:

The answer is 1823.9

Explanation:

Solution

Given that:

m = 5.5 kg/s

= m₁ = m₂ = m₃

The work carried out by the energy balance is given as follows:

m₁h₁ = m₂h₂ +m₃h₃ + w

Now,

By applying the steam table we have that<

p₃ = 50 kPa

T₃ = 100°C

Which is

h₃ = 2682.4 kJ/KJ

s₃ = 7.6953 kJ/kgK

Since it is an isentropic process:

Then,

p₂ =  500 kPa

s₂=s₃ = 7.6953 kJ/kgK

which is

h₂ =3207.21 kJ/KgK

p₁ = 3HP0

s₁ = s₂=s₃ = 7.6953 kJ/kgK

h₁ =3854.85 kJ/kg

Thus,

Since 5 % of this flow diverted to p₂ =  500 kPa

Then

w =m (h₁-0.05 h₂ -0.95 )h₃

5.5(3854.85 - 0.05 * 3207.21  - 0.95 * 2682.4)

5.5( 3854.83 * 3207.21 - 0.95 * 2682.4)

5.5 ( 123363249.32 -0.95 * 2682.4)

w=1823.9

Follow traffic signs , Keep distance between cars , Be patient in traffic.

Answer:

The average thickness of the blubber is 0.077 m

Explanation:

Here, we want to calculate the average thickness of the Walrus blubber.

We employ a mathematical formula to calculate this;

The rate of heat transfer(H) through the Walrus blubber = dQ/dT = KA(T2-T1)/L

Where dQ is the change in amount of heat transferred

dT is the temperature gradient(change in temperature) i.e T2-T1

dQ/dT = 220 W

K is the conductivity of fatty tissue without blood = 0.20 (J/s · m · °C)

A is the surface area which is 2.23 m^2

T2 = 37.0 °C

T1 = -1.0 °C

L is ?

We can rewrite the equation in terms of L as follows;

L × dQ/dT = KA(T2-T1)

L = KA(T2-T1) ÷ dQ/dT

Imputing the values listed above;

L = (0.2 * 2.23)(37-(-1))/220

L = (0.2 * 2.23 * 38)/220 = 16.948/220 = 0.077 m

Answer:

the overall elongation δ of the bar is  1.2337 mm

Explanation:

From the information given :

According to the principle of superposition being applied to the axial load P of the system; we have:

[tex]\delta = \delta_{AB} +\delta_{BC} + \delta_{CD}[/tex]    

where;

δ = overall elongation

[tex]\delta _{AB}[/tex] = elongation of bar AB

[tex]\delta _{BC}[/tex] = elongation of  bar BC

[tex]\delta _{CD} =[/tex]  elongation of bar CD]

If we replace; [tex]\dfrac{PL}{AE}[/tex] for  δ  and bt for area;

we have:

[tex]\delta = \dfrac{P_{AB}L_{AB}}{(b_{AB}t)E} +\dfrac{P_{BC}L_{BC}}{(b_{BC}t)E}+\dfrac{P_{CD}L_{CD}}{(b_{CD}t)E}[/tex]

where ;

P = load

L = length of the bar

A = area of the cross-section

E = young modulus of elasticity

Let once again replace:

P for [tex]P_{AB}, P_{BC} , P_{CD}[/tex]  (since load in all member of AB, BC and CD will remain the same )

[tex]\dfrac{L}{4}[/tex] for [tex]L_{AB}[/tex],  

[tex]\dfrac{L}{2}[/tex] for [tex]L_{BC}[/tex] and

[tex]\dfrac{L}{4}[/tex] for [tex]L_{CD}[/tex]

[tex]2\dfrac{b}{3}[/tex] for  [tex]b_{BC}[/tex]

b for  [tex]b_{CD}[/tex]

[tex]\delta = \dfrac{P (\dfrac{L}{4})}{btE}+ \dfrac{P (\dfrac{L}{2})}{2 \dfrac{b}{3}tE}+\dfrac{P (\dfrac{L}{4})}{btE}[/tex]

[tex]\delta = \dfrac{PL}{btE}[\dfrac{1}{4}+ \dfrac{1}{2}*\dfrac{3}{2}+ \dfrac{1}{4}][/tex]

[tex]\delta = \dfrac{5}{4}\dfrac{PL}{btE} --- \ (1)[/tex]

The stress in the central portion can be calculated as:

[tex]\sigma = \dfrac{P}{A}[/tex]

[tex]\sigma = \dfrac{P}{\dfrac{2}{3}bt}[/tex]

[tex]\sigma = \dfrac{3P}{2bt}[/tex]

So; Now:

[tex]\delta = \dfrac{5}{4}* \dfrac{2 * \sigma}{3}*\dfrac{L}{E}[/tex]

[tex]\delta= \dfrac{5}{4}* \dfrac{2 * 200}{3}*\dfrac{570}{77*10^3 \ MPa}[/tex]

δ = 1.2337 mm

Therefore, the overall elongation δ of the bar is  1.2337 mm

Answer:

None of these

Explanation:

There are different types of amplifiers, and each has different characteristics.

Therefore, the correct answer is "None of these "

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:

a = 0.07m or 70mm

b = 0.205m or 205mm

Explanation:

Given the following data;

Modulus of rigidity, G = 14MPa=14000000Pa.

c = 80mm = 0.08m.

P = 46kN=46000N.

Shearing stress (r) in the rubber shouldn't exceed 1.4MPa=1400000Pa.

Deflection (d) of the plate is to be at least 7mm = 0.007m.

From shearing strain;

[

[tex]Modulus Of Elasticity, E = \frac{d}{a} =\frac{r}{G}[/tex]

Making a the subject formula;

[tex]a = \frac{Gd}{r}[/tex]

Substituting into the above formula;

[tex]a = \frac{14000000*0.007}{1400000}[/tex]

[tex]a = \frac{98000}{1400000}[/tex]

[tex]a = 0.07m or 70mm[/tex]

a = 0.07m or 70mm.

Also, shearing stress;

[tex]r = \frac{P}{2bc}[/tex]

Making b the subject formula;

[tex]b = \frac{P}{2cr}[/tex]

Substituting into the above equation;

[tex]b = \frac{46000}{2*0.08*1400000}[/tex]

[tex]b = \frac{46000}{224000}[/tex]

[tex]b = 0.205m or 205mm[/tex]

b = 0.205m or 205mm

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:

Explanation:

The image attached to the question is shown in the first diagram below.

From the diagram given ; we can deduce a free body diagram which will aid us in solving the question.

IF we take a look at the second diagram attached below ; we will have a clear understanding of what the free body diagram of the system looks like :

From the diagram; we can determine the length of BC by using pyhtagoras theorem;

SO;

[tex]L_{BC}^2 = L_{AB}^2 + L_{AC}^2[/tex]

[tex]L_{BC}^2 = (3.5+2.5)^2+ 4^2[/tex]

[tex]L_{BC}= \sqrt{(6)^2+ 4^2}[/tex]

[tex]L_{BC}= \sqrt{36+ 16}[/tex]

[tex]L_{BC}= \sqrt{52}[/tex]

[tex]L_{BC}= 7.2111 \ m[/tex]

The cross -sectional of the cable is calculated by the formula :

[tex]A = \dfrac{\pi}{4}d^2[/tex]

where d = 4mm

[tex]A = \dfrac{\pi}{4}(4 \ mm * \dfrac{1 \ m}{1000 \ mm})^2[/tex]

A = 1.26 × 10⁻⁵ m²

However, looking at the maximum deflection  in length [tex]\delta[/tex] ; we can calculate for the force [tex]F_{BC[/tex] by using the formula:

[tex]\delta = \dfrac{F_{BC}L_{BC}}{AE}[/tex]

[tex]F_{BC} = \dfrac{ AE \ \delta}{L_{BC}}[/tex]

where ;

E = modulus elasticity

[tex]L_{BC}[/tex] = length of the cable

Replacing 1.26 × 10⁻⁵ m² for A; 200 × 10⁹ Pa for E ; 7.2111 m for [tex]L_{BC}[/tex] and 0.006 m for [tex]\delta[/tex] ; we have:

[tex]F_{BC} = \dfrac{1.26*10^{-5}*200*10^9*0.006}{7.2111}[/tex]

[tex]F_{BC} = 2096.76 \ N \\ \\ F_{BC} = 2.09676 \ kN[/tex]     ---- (1)

Similarly; we can determine the force [tex]F_{BC}[/tex] using the allowable  maximum stress; we have the following relation,

[tex]\sigma = \dfrac{F_{BC}}{A}[/tex]

[tex]{F_{BC}}= {A}*\sigma[/tex]

where;

[tex]\sigma =[/tex] maximum allowable stress

Replacing 190 × 10⁶ Pa for [tex]\sigma[/tex] ; we have :

[tex]{F_{BC}}= 1.26*10^{-5} * 190*10^{6} \\ \\ {F_{BC}}=2394 \ N \\ \\ {F_{BC}}= 2.394 \ kN[/tex]     ------ (2)

Comparing (1) and  (2)

The magnitude of the force [tex]F_{BC} = 2.09676 \ kN[/tex] since the elongation of the cable should not exceed 6mm

Finally applying the moment equilibrium condition about point A

[tex]\sum M_A = 0[/tex]

[tex]3.5 P - (6) ( \dfrac{4}{7.2111}F_{BC}) = 0[/tex]

[tex]3.5 P - 3.328 F_{BC} = 0[/tex]

[tex]3.5 P = 3.328 F_{BC}[/tex]

[tex]3.5 P = 3.328 *2.09676 \ kN[/tex]

[tex]P =\dfrac{ 3.328 *2.09676 \ kN}{3.5 }[/tex]

P = 1.9937 kN

Hence; the maximum load P that can be applied is 1.9937 kN

Find the given attachments for complete explanation

Answer:

Overall project duration

Explanation:

Scheduling can best be defined as the process used to determine a overall project duration.

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:

The frequency components in the message signal are

f1 = 100Hz, f2 = 200Hz and f3 = 400Hz

When amplitude modulated with a carrier signal of frequency fc = 100kHz

Generates the following frequency components

Lower side band

[tex]100k - 100 = 99.9kHz\\\\100k - 200 = 99.8kHz\\\\100k - 400 = 99.6kHz\\\\[/tex]

Carrier frequency 100kHz

Upper side band

[tex]100k + 100 = 100.1kHz\\\\100k + 200 = 100.2kHz\\\\100k + 400 = 100.4kHz[/tex]

After passing through the SSB filter that filters the lower side band, the transmitted frequency component will be

[tex]100k, 100.1k, 100.2k\ \texttt {and}\ 100.4kHz[/tex]

At the receive these are mixed (superheterodyned) with local ocillator frequency whichh is 100.02KHz, the output frequencies will be

[tex]100.02 - 100.1k = 0.08k = 80Hz\\\\100.02 - 100.2k = 0.18k = 180Hz\\\\100.02 - 100.4 = 0.38k = 380Hz[/tex]

After passing through the SSB filter that filters the higher side band, the transmitted frequency component will be

[tex]100k, 99.9k, 99.8k\ \ and \ \99.6kHz[/tex]

At the receive these are mixed (superheterodyned) with local oscillator frequency which is 100.02KHz, and then fed to the detector whose output frequencies will be

[tex]100.02 - 99.9k = 0.12k = 120Hz\\\\100.02 - 99.8k = 0.22k = 220Hz\\\\100.02 - 99.6k = 0.42k = 420Hz[/tex]

A) The frequency Components of the Detector Output are;

80 Hz, 120 Hz and 380 Hz

B) The frequency Components if only the lower sideband is transmitted are; 120 Hz, 220 Hz and 420 Hz

A) We are given the frequency components in the message signal as;

We are told that the carrier signal has a frequency; fc = 100kHz

Thus, the frequency components generated are;

Lower side band:

Upper side band:

We are told that the local oscillator now supplies a sinusoidal wave of frequency 100.02 kHz.

Thus, the output frequencies are;

100.02 kHz - 100.1 kHz = 80 Hz

100.02 kHz - 100.2 kHz = 180 Hz

100.02 kHz - 100.4 kHz = 380 Hz

B) Repeating the analysis assuming only the lower sideband is repeated gives us the frequencies as;

100.02 kHz - 99.9 kHz = 120 Hz

100.02 kHz - 99.8 kHz = 220 Hz

100.02 kHz - 99.6 kHz = 420 Hz

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