The propeller shaft of the submarine experiences both torsional and axial loads. Draw Mohr's Circle for a stress element on the outside surface of the solid shaft. Determine the principal stresses, the maximum in-plane shear stress and average normal stress using Mohr's Circle.

Answers

Answer 1

Answer: Attached below is the missing detail and Mohr's circle.

i) б1 =  9.6 Ksi

б2 = -10.7 ksi

ii) 10.2 Ksi

iii)  -0.51Ksi

Explanation:

First step :

direct compressive stress on shaft

бd = P / π/4 * d^2

      = -20 / 0.785 * 5^2  = -1.09 Ksi

shear stress at the outer surface due to torsion

ζ = 16*T / πd^3

  = (16 * 250 ) / π * 5^3  = 010.19 Ksi

Calculate the Principal stress, maximum in-plane shear stress and average normal stress

Using Mohr's circle ( attached below )

i) principal stresses:

б1 = 4.8 cm * 2 = 9.6 Ksi

б2 = -5.35 cm * 2 = -10.7 ksi

ii) maximum in-plane shear stress

ζ  = radius of Mohr's circle

   = 5.1 cm = 10.2 Ksi   ( Given that ; 1 cm = 2Ksi )

iii) average normal stress

 = 9.6 + ( - 10.7 ) / 2

  = -0.51Ksi

The Propeller Shaft Of The Submarine Experiences Both Torsional And Axial Loads. Draw Mohr's Circle For
The Propeller Shaft Of The Submarine Experiences Both Torsional And Axial Loads. Draw Mohr's Circle For

Related Questions

what are qualifications that are required to be an architect​

Answers

Answer:

Bachelor's degree

In order to become a licensed architect in the US and the District of Columbia, applicants are required to complete a professional degree in architecture, gain on-the-job experience through a paid internship.

How do your arm muscles work to lift a mug of coffee to your lips?
A.
The biceps muscle contracts while the triceps muscle relaxes. Only then can the forearm move up to lift the mug.
B.
The muscle tendons in your arms stretch to their limit as a result of the flexibility exercises that lift the mug.
C.
Your muscles rapidly increase the speed at which they twitch to trigger your arm movement.
D.
Your biceps and triceps muscles first relax, and then the arm muscle raises the mug.

Answers

Answer:

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

Explanation:

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

Generally, skeletal muscles are only found in vertebrates.

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

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

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

To bend the elbow and raise the forearm, the biceps muscle contracts and the triceps muscle relaxes.To straighten the elbow and lower the forearm, the biceps muscle relaxes and the triceps muscle contracts.

Answer:A

Explanation:

got it right on plato

Set the leak rate to zero and choose a non-zero value for the proportional feedback gain.Restart the simulation and turn on the outflow valve.What happens to the liquid level in the tank?Repeat this process with higher and lower values for the proportional feedback gain.What happens when the proportional feedback gain is increased?What happens when it is decreased?Find the proportional gain that will reach steady state the quickest without oscillationin the state of the valve and restart the simulation.What is the system time constant, as determined from the tank level versus time plot.

Answers

Answer:

Explanation:

The proportional gain K is usually a fixed property of the controller . If proportional gain is increased , The sensitivity of the controller to error is increased but the stability is impaired. The system approaches the behaviour of on off controlled system and it response become oscillatory

A hollow pipe is submerged in a stream of water so that the length of the pipe is parallel to the velocity of the water. If the water speed doubles and the cross-sectional area of the pipe triples, what happens to the volume flow rate of the water passing through it?

Answers

Answer:

increases by a factor of 6.

Explanation:

Let us assume that the initial cross sectional area of the pipe is A m² while the initial velocity of the water is V m/s², hence the flow rate of the water is:

Initial flow rate = area * velocity = A * V = AV m³/s

The water speed doubles (2V m/s) and the cross-sectional area of the pipe triples (3A m²), hence the volume flow rate becomes:

Final flow rate = 2V * 3A = 6AV m³/s = 6 * initial flow rate

Hence, the volume flow rate of the water passing through it increases by a factor of 6.

A container can be made from steel [β = 36 × 10-6 (C°)-1] or lead [β = 87 × 10-6 (C°)-1]. A liquid is poured into the container, filling it to the brim. The liquid is either water [β = 207 × 10-6 (C°)-1] or ethyl alcohol [β = 1120 × 10-6 (C°)-1]. When the full container is heated, some liquid spills out. To keep the overflow to a minimum, from what should the container be made and what should the liquid be

Answers

Answer:

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

Explanation:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

A work-mode-choice model is developed from data acquired in the field in order to determine the probabilities of individual travelers selecting various modes. The mode choices include automobile drive-alone (DL), automobile shared-ride (SR), and bus (B). The utility functions are estimated as follows:
UDL = 2.6 - 0.3 (costDL) - 0.02 (travel timeDL)
USR = 0.7 - 0.3 (costSR) - 0.04 (travel timeSR)
UB = -0.3 (costB) - 0.01 (travel timeB)
where cost is in dollars and time is in minutes. The cost of driving an automobile is exist5.50 with a travel time of 21 minutes, while the bus fare is exist1.25 with a travel time of 27 minutes. How many people will use the shared-ride mode from a community of 4500 workers, assuming the shared-ride option always consists of three individuals sharing costs equally?
a. 314
b. 828
c. 866
d. 2805

Answers

Answer:

b. 828

Explanation:

UDL = 2.6 - 0.3 [5.5] - 0.02 [ 21 ] = 0.53

USR = 0.7 - 0.3 [5.5 / 2 ] - 0.04 [ 21 ] = -0.69

UB = -0.3 [ 1.25 ] - 0.01 = -0.645

Psr = [tex]\frac{e^{-0.53} }{e^{-0.53} + e^{-0.69} + e^{-0.645} }[/tex]

Solving the equation we get 0.184.

Number of people who will take shared ride is:

0.184 * 4500 = 828 approximately.

All of the following safety tips are true EXCEPT Select one: a. It is not acceptable to handle broken glass with your bare hands b. It is acceptable to grasp the electrical cord when removing an electrical plug from its socket c. It is not acceptable to immerse hot glassware in cold water d. It is not acceptable to reuse dirty glassware

Answers

Answer:

Explanation:

B. you would grab the plug closest to the outlet

If you don't have enough experience, it's always best to leave socket changing to the experts. If you make a mistake, you might inflict harm and potentially endanger yourself and other people. Read on if you're interested in learning how to change a socket safely. Thus, option D is correct.

What, removing an electrical plug from its socket?

Grip the plug, not the electrical cable, when taking an electrical plug out of its socket. Before handling an electrical switch, socket, or outlet, hands must be fully dry.

Reduce the extra so that it rests only on top of the existing plasterboard. If necessary, push it back a little by using your finger. Fill the dent with ready-mixed filler or powdered filler, whichever you want, and bring it flush with the surrounding wall. Allow to dry, then sand off any excess.

Therefore, It is acceptable to grasp the electrical cord when removing an electrical plug from its socket

Learn more about electrical plug here:

https://brainly.com/question/28932892

#SPJ5

Consider two houses that are identical, except that the walls are built using bricks in one house, and wood in the other. The walls of the brick house are twice as thick. Which house do you think will be more energy efficient?

Answers

Answer:

Walls Built Using Bricks and Wood

The brick house is more energy-efficient than the one built with wood.

Explanation:

Because of their high thermal mass, which gives bricks the ability to absorb heat and release it over time, bricks remain more energy-efficient than other building materials, including wood.  In summer, bricks leave your home cool.  In winter, they make it warm.  With these two advantages provided by bricks over other building materials, bricks are the most energy-efficient building material.

Water is pumped steadily through a 0.10-m diameter pipe from one closed pressurized tank to another tank. The pump adds 4.0 kW of energy to the water and the head loss of the flow is 10 m. Determine the velocity of the water leaving the pump and discharging into tank B.

Answers

Complete Question

Complete Question is attached below.

Answer:

[tex]V'=5m/s[/tex]

Explanation:

From the question we are told that:

Diameter [tex]d=0.10m[/tex]

Power [tex]P=4.0kW[/tex]

Head loss [tex]\mu=10m[/tex]

 [tex]\frac{P_1}{\rho g}+\frac{V_1^2}{2g}+Z_1+H_m=\frac{P_2}{\rho g}+\frac{V_2^2}{2g}+Z_2+\mu[/tex]

 [tex]\frac{300*10^3}{\rho g}+35+Hm=\frac{500*10^3}{\rho g}+15+10[/tex]

 [tex]H_m=(\frac{200*10^3}{1000*9.8}-10)[/tex]

 [tex]H_m=10.39m[/tex]

Generally the equation for Power is mathematically given by

 [tex]P=\rho gQH_m[/tex]

Therefore

 [tex]Q=\frac{P}{\rho g H_m}[/tex]

 [tex]Q=\frac{4*10^4}{1000*9.81*10.9}[/tex]

 [tex]Q=0.03935m^3/sec[/tex]

Since

 [tex]Q=AV'[/tex]

Where

 [tex]A=\pi r^2\\A=3.142 (0.05)^2[/tex]

 [tex]A=7.85*10^{-3}[/tex]

Therefore

 [tex]V'=\frac{0.03935m^3/sec}{7.85*10^{-3}}[/tex]

 [tex]V'=5m/s[/tex]

A signalized intersection has a sum of critical flow ratios of 0.72 and a total cycle lost time of 12 seconds. Assuming a critical intersection v/c ration of 0.9, calculate the minimum necessary cycle length.

Answers

Answer:

[tex]T_o=82.1sec[/tex]

Explanation:

From the question we are told that:

Lost Time [tex]t=12secs[/tex]

Sum of critical flow ratios [tex]X=0.72[/tex]

Generally the Webster Method's equation for Optimum cycle time is is mathematically given by

 [tex]T_o=\frac{1.5t+5}{1-x}[/tex]

 [tex]T_o=\frac{1.5*12+5}{1-0.72}[/tex]

 [tex]T_o=82.1sec[/tex]

A 0.82-in-diameter aluminum rod is 5.5 ft long and carries a load of 3000 lbf. Find the tensile stress, the total deformation, the unit strains, and the change in the rod diameter.

Answers

Answer:

Tensile stress = 0.1855Kpsi

Total deformation = 0.0012243 in

Unit strain =  1.855 *10^-5   or  18.55μ

Change in the rod diameter = 5.02 * 10^ -6 in

Explanation:

Data given: D= 0.82 in

                   L = 5.5 ft * 12 = 66 in

load (p) = 3000 (Ibf) /32.174 = 93.243 Ibm

Area = (π/4) D² = (π/4) 0.82²  = 0.502655 in²

∴ Tensile stress Rt = P/A = 93.243/0.502655 = 185.50099 pound/in²

                           Rt = 0.1855 Kpsi

∴ Total deformation = PL / AE = Rt * L/ Eal

                                 = 0.1855 * 10³  * 66 / 10000 * 10³

                                 = 0.0012243 in

∴the unit strains = total deformation / L = 0.0012243/ 66

                          =0.00001855 = 1.855 *10^-5

                         = 18.55μ

∴ Change in rod   Δd/ d = μ ΔL/L

                           = (0.33) 1.855 *10^-5 * 0.82

                           = 5.02 * 10^ -6 in

6. When the engine stalls or the power unit fails, on a car with power
brakes, the service brake pedal will
A. Take about the same amount of pressure
B. Take more pressure to stop
C. Take less pressure to stop
D. Become locked in place and no longer help stop the car

Answers

B

But

I think

So yea it prob isn’t

An ideal neon sign transformer provides 9130 V at 51.0 mA with an input voltage of 240 V. Calculate the transformer's input power and current.

Answers

Answer:

Input power = 465.63 W

current = 1.94 A

Explanation:

we have the following data to answer this question

V = 9130

i = 0.051

the input power = VI

I = 51.0 mA = 0.051

= 9130 * 0.051

= 465.63 watts

the current = 465.63/240

= 1.94A

therefore the input power is 465.63 wwatts

while the current is 1.94A

the input power is the same thing as the output power.

it is a small sharp and printed item for fine worker in trimming scallops clipping threads and cutting large eyelets​

Answers

Answer:

embroidery scissor

Explanation:

is small, sharp and pointed good for fine work use trimming scallops,clipping threads,and cutting large eyelets.

hope this helps

Consider the titration of 100.0 mL of 0.200 M CH3NH2 by 0.100 M HCl.
For each volume of HCl added, decide which of the components is a major species after the HCl has reacted completely.
Kb for CH3NH2 = 4.4 x 10-4.
What species are present based off what is being added?
200.00 mL HCl added
yes no H+
yes no H2O
yes no Cl-
yes no CH3NH2
yes no CH3NH3+
300.00 mL HCl added
yes no H+
yes no H2O
yes no Cl-
yes no CH3NH2
yes no CH3NH3+
Calculate the pH at the equivalence point for this titration?

Answers

Answer:

The answers are in the explanation. The pH is 5.91

Explanation:

The CH3NH2 reacts with HCl as follows:

CH3NH2 + HCl → CH3NH3⁺ + Cl⁻

When 200mL of HCl are added, the moles of CH3NH2 and HCl are reacting completely producing CH3NH3+ and Cl-. That means the species present are:

no H+. All reacted

yes H2O. Because the water is present in the solutions of HCl and CH3NH2

yes Cl-. Is a product of the reaction

Yes CH3NH2. Is consumed in the reaction but comes from the equilibrium of CH3NH3+

yes CH3NH3+. Is the other product of the reaction. MAJOR SPECIES

When 300.00mL of HCl are added, 100mL are in excess:

yes H+. Is in excess: H+ + Cl- = HCl in water. MAJOR SPECIES. Determine the pH of the solution.

yes H2O. Is present because the reactants are diluted.  

yes Cl-. Is a product of reaction and comes from HCl.

Yes CH3NH2. The reactant is over but comes from the equilibrium of CH3NH3+

yes no CH3NH3+. Yes. Is a product and remains despite HCl is in excess.

To find the pH:

At equivalence point the ion that determines pH is CH3NH3+. Its concentration is:

0.100L * (0.200mol/L) = 0.0200 moles / 0.300L = 0.0667M CH3NH3+

The equilibrium of CH3NH3+ is:

Ka = Kw/kb = 1x10-14/4.4x10-4 = 2.273x10-11 = [H+] [CH3NH2] / [CH3NH3+]

As both [H+] [CH3NH2] comes from the same equilibrium:

[H+] =  [CH3NH2] = X

2.273x10-11 = [X] [X] / [0.0667M]

1.5159x10-12 = X²

X = 1.23x10-6M = [H+]

As pH = -log [H+]

pH = 5.91

The pH at the equivalent point for this titration is "5.91".

pH Calculation:

[tex]CH_3NH_2 = 0.200\ M\\\\ \text{volume} = 100.0\ mL = 0.100\ L\\\\HCl = 0.100\ M\\\\[/tex]

We must now quantify the pH well at the equivalence point.

We know that even at the point of equivalence, moles of acid and moles of the base are equivalent. As such, first, we must calculate the number of moles of the given base.

Calculating the Moles in [tex]CH_3NH_2 = 0.200\ M \times 0.100\ L = 0.0200\ moles[/tex]

Calculating the Moles in [tex]HCl = 0.0200 \ moles[/tex]

Calculating the volume of [tex]HCl[/tex]:

[tex]\to \text{Molarity} = \frac{ \text{moles}}{\text{volume \ (L)}} \\\\\to \text{Volume} = \frac{\text{moles}}{\text{molarity}}\\\\[/tex]

                [tex]= \frac{0.0200 \ moles}{ 0.100\ M}\\\\= 0.200 \ L\\\\= 200 \ mL\\\\[/tex]

Calculating the reaction among the acid and base:

[tex]CH_3NH_2 + HCl \longrightarrow CH_3NH_3^{+} + Cl^-[/tex]

[tex]0.0200 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.0200 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.0200[/tex]

Therefore the conjugate acid of the bases exists at the standard solution.

Then we must calculate the new molar mass of [tex]CH_3NH_3^+[/tex].

Total volume[tex]= 100 + 200 = 300\ mL = 0.300\ L[/tex]

[tex][CH_3NH_3^+] = \frac{0.0200\ mole}{ 0.300\ L}= 0.0667\ M[/tex]

Using the ICE table

[tex]CH_3NH_3^+ + H_2O \longrightarrow CH_3NH_2 + H_3O^+[/tex]

[tex]I \ \ \ \ \ \ \ \ \ \ 0.0667 \ \ \ \ \ \ \ \ \ 0\ \ \ \ \ \ \ \ \ 0\\\\C\ \ \ \ \ \ \ \ -x\ \ \ \ \ \ \ \ +x \ \ \ \ \ \ \ \ +x\\\\E \ \ \ \ \ \ \ \ \ \ \ \ 0.0667-x \ \ \ \ \ \ \ \ \ \ \ \ +x \ \ \ \ \ \ \ \ \ \ \ \+x\\\\\to Ka = \frac{[CH_3NH_2] [H_3O^+] }{[CH_3NH_3^+]}[/tex]

Calculating [tex]K_a[/tex] from [tex]K_b[/tex]

[tex]\to K_a \times K_b = 1\times 10^{-14}\\\\\to K_a = \frac{1\times 10^{-14}}{4.4\times 10^{-4}} = 2.27\times 10^{-11}\\\\[/tex]

                           [tex]= 2.27\times 10^{-11} \\\\= x\times \frac{x}{(0.0667-x)}[/tex]

The x in the 0.0667-x can be ignored since the Ka value is just too small and it also does not follow the five percent criteria.

[tex]\to 2.27 \times 10^{-11} \times 0.0667 = x_2\\\\\to x_2 = 1.515\times 10^{-12}\\\\\to x = 1.23\times 10^{-6}\ M\\\\\to [H_3O^+] = x = 1.23\times 10^{-6}\ M\\\\[/tex]

We have the formula to calculate pH.

[tex]\to pH = - \log [H_3O^+] = - \log 1.23\times 10^{-6}\ M= 5.91[/tex]

The pH at the equivalent point for this titration is "5.91".

Find out more information about the pH here:

brainly.com/question/15289741

) Please label the following statements as either True (T) or False (F). (a) In general, the greater the % of cold work, the smaller the recrystallization grain size. (b) The higher the annealing temperature, the smaller the recrystallization grain size. (c) The greater the % of cold work, the lower the recrystallization temperature.

Answers

Answer:

A. This option is true

B. This option is false

C. This option is true

Explanation:

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

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

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

Blocks A and B each have a mass m. Determine the largest horizontal force P which can be applied to B so that A will not move relative to B. All surfaces are smooth.

Answers

Answer:

The answer is "15 N".

Explanation:

Please find the complete question in the attached file.

In frame B:

For just slipping:

[tex]\to \frac{P}{2} \cos \theta =mg \sin \theta\\\\\to P=2 mg \tan \theta \\\\[/tex]

        [tex]=2 \times 1 \times g \times \tan 37^{\circ}\\\\ =2 \times 10 \times \frac{3}{4}\\\\ =15 \ N[/tex]

An industrial load with an operating voltage of 480/0° V is connected to the power system. The load absorbs 120 kW with a lagging power factor of 0.77. Determine the size of the capacitor in vars that is necessary to raise the power factor to 0.9 lagging.

Answers

Answer:

[tex]Q=41.33 KVAR\ \\at\\\ 480 Vrms[/tex]

Explanation:

From the question we are told that:

Voltage [tex]V=480/0 \textdegree V[/tex]

Power [tex]P=120kW[/tex]

Initial Power factor [tex]p.f_1=0.77 lagging[/tex]

Final Power factor [tex]p.f_2=0.9 lagging[/tex]

Generally the equation for Reactive Power is mathematically given by

Q=P(tan \theta_2-tan \theta_1)

Since

[tex]p.f_1=0.77[/tex]

[tex]cos \theta_1 =0.77[/tex]

[tex]\theta_1=cos^{-1}0.77[/tex]

[tex]\theta_1=39.65 \textdegree[/tex]

And

[tex]p.f_2=0.9[/tex]

[tex]cos \theta_2 =0.9[/tex]

[tex]\theta_2=cos^{-1}0.9[/tex]

[tex]\theta_2=25.84 \textdegree[/tex]

Therefore

[tex]Q=P(tan 25.84 \textdegree-tan 39.65 \textdegree)[/tex]

[tex]Q=120*10^3(tan 25.84 \textdegree-tan 39.65 \textdegree)[/tex]

[tex]Q=-41.33VAR[/tex]

Therefore

The size of the capacitor in vars that is necessary to raise the power factor to 0.9 lagging is

[tex]Q=41.33 KVAR\ \\at\\\ 480 Vrms[/tex]

Imagine a cantilever beam fixed at one end with a mass = m and a length = L. If this beam is subject to an inertial force and a uniformly distributed load = w, what is the moment present at a length of L/4?

Answers

Answer:

jsow

hfhcffnbxhdhdhdhdhdhdddhdhdgdhdhdhdhdhdhhhdhdjsksmalalaksjdhfgrgubfghhhhhhh

Explanation:

j

grudb

Technician A says that a graphing multi-meter may be used to verify signals going to and from electrical and electronic components. Technician B says that digital storage oscilloscope may be used to verify signals going to and from electrical and electronic components. Who is correct

Answers

Answer:

Both are correct.

Explanation:

Graphing multi meter is used to verify signals that move from electrical components. Digital oscilloscope is an equipment which stores and analyzes input signals with digital technique.

In an international film festival, a penal of 11 judges is formed to judge the best film. At
last two films FA and FB were considered to be the best where the opinion of judges got
divided. Six judges where in favor of FA whereas five in favor of FB. A random sample
of five judges was drawn from the panel. Find the probability that out of five judges,
three are in favor of film FA.Enunciate demerits of classical probability.

Answers

Answer:

International Film Festival

Judging the best best film:

a. The probability that out of five judges (random sample),  three are in favor of film FA is:

= 33%.

b. The demerits of classical probability are:

1. Classical probability can only be used with events that have definite numbers of possible outcomes.  

2. Classical probability can only handle events where each outcome is equally likely.

3. Classical probability is based on the assumption of linear relationship (which is not always true in real life) between the latent variable and observed scores.

Explanation:

a) Number of judges = 11

Number of judges in favor of FA film = 6

Number of judges in favor of FB film = 5

Probability of judges in favor of FA film = 6/11

Probability of judges in favor of FB film = 5/11

Random sample of judges = 5

Probability that out of five judges, three are in favor of film FA = 3/5 * 6/11

= 18/55

= 33%

b) Classical probability is the simple probability showing that each event has equal chance of happening.  It can be contrasted with empirical probability that is obtained from experiments.

Alice and Bob both have RSA Public-Private key pairs: (PUA, PRA) and (PUB, PRB). They also have cryptographic functions E_AES / D_AES to encrypt / decrypt using AES; and E_RSA and D_RSA to encrypt / decrypt using RSA. Alice wants to sent a high resolution video of a large secret facility to Bob.
A. Show how Alice can securely and efficiently send the video to Bob. You are required to use the cryptographic functions above to get full credit;
B. Does your solutions assure confidentiality? How / Why not?
C. Does your solutions assure non-repudiation? How / Why not?
D. Does your solutions assure integrity? How / Why not?
E. Does your solutions assure replay attacks? How / Why not?

Answers

Solution :

B. yes, the given solution assures confidentiality. The sender Alice encrypting his messages with its own private key PRA which provides authentication. Sender Alice further encrypts his messages with the receiver's public key PUB provides confidentiality.

C. So the given solution provides non repudiation. Alice and Bob who are exchanging messages. In one case, Alice denies sending a messages to Bob that he claims to have received being able to counter Alice's denial is caused non repudiation of origin.

D. The given solution provides integrity. Because it provides authentication and have not been changed.

E. It does not provide replay attacks because it does not captures the traffic. The client does not receive the messages twice.

Given resistance 30ohms Inductance 200mH is connected to a 230v,50hZ supply. Impedance 69.6ohms Calculate current consumed?

Answers

Answer:

the current consumed is 3.3 A

Explanation:

Given;

resistance, R = 30 ohms

inductance, L = 200 mH

Voltage supply, V = 230 V

frequency of the coil, f = 50 Hz

impedance, Z = 69.6 Ohms

The current consumed is calculated as;

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

Therefore, the current consumed is 3.3 A

Explain by Research how a basic generator works ? using diagram

Answers

Correcto no se muy bien de que se trata el tema porque está en inglés.
Sorry

1. A manufacturing cell with two workers is responsible for producing a small frying pan with a required takt time of 496 seconds. The material passes through two processes: a deep drawing process and a trimming process. The average cycle time for the deep drawing process is 450 seconds and average cycle time for trimming is 430 seconds. (2 pts.)
a. Does the work cell have adequate capacity to meet demand? Explain.
b. What is the required daily production capacity of the work cell (in number of frying pans per day)? Assume 480 minutes/workday of available time.
2. What is the total daily idle time for both workers in Problem 1? Report your answer in (a) seconds of idle time and (b) as a percentage of total working time for the cell. (2 pts.)

Answers

Answer:

Explanation:

[tex]496=\frac{480\times 60}{demand}[/tex]

demand per day = 58 pans

Due to availability of two workers we can have parallel we can have deep drawing and trimming operations simultaneously.

Hence the cycle time would be the greater time of the two operations.

cycle time = 450 seconds

[tex]\text{capacity of work cell}=\frac{\text{available working time}}{\text{cycle time}}[/tex]

[tex]\text{capacity of work cell}=\frac{480\times 60}{450}[/tex]

[tex]\text{capacity of work cell}=64 ~pans[/tex] (which is greater than the demand of 58 pans)

Therefore the work cell has sufficient capacity and time (496 sec.>cycle time 450 sec) to meet the demand.

b)

Required daily production is 58 pans

A designer needs to select the material for a plate under tensile stress. Assuming that the applied tensile force is 13,000 lb and the area under the stress is 4 square inches, determine which material should be selected to assure safety. Assume safety factor is 2. Material A: Ultimate Tensile stress is 8000 lb/in2Material B: Ultimate Tensile stress is 5500 lb/in2

Answers

Wow Muy bien no se que dice pero bien.

where are the field poles mounted on an alternator

Answers

Answer:

The magnetic field for this type of alternator is established by a set of stationary field poles mounted on the periphery of the alternator frame. The field flux created by these poles is cut by conductors inserted in slots on the surface of the rotating armature.

The following measurements are taken on particular junction diodes for which V is the terminal voltage and I is the diode current. For each diode, estimate values of Is and the terminal voltage at 10% of the measured current.
(a) V = 0.700 V at I = 1.00 A.
(b) V = 0.650 V at I = 1.00 mA.
(c) V = 0.650 V at I = 10 mu A.
(d) V = 0.700V at I = 100 mA.

Answers

Poop Neal sbskqlgnwnf

The values of Is and V are as: (a) [tex]Is = 2.34 \times 10^{-11} A[/tex] and V = 0.581 V. (b) [tex]Is = 4.56 \times 10^{-15} A[/tex] and V = 0.516 V. (c) [tex]Is = 1.18 \times 10^{-16} A\\[/tex] and V = 0.459 V. (d) [tex]Is = 2.34 \times 10^{-11} A[/tex] and V = 0.581 V.

The relation between the current and voltage of a diode is given by the Shockley diode equation. It is an exponential function and can be given by the following equation:

[tex]I = Is \times (e^{V/Vt} - 1)[/tex]

Where

I = currentV = voltageVt = thermal voltageIs = reverse saturation current.

(a)

Given that:

V = 0.700 V

And I = 1.00 A.

Substituting these values in the equation above to get,

[tex]1.00 A = Is \times (e^{0.700 V / 0.025 V} - 1)\\Is = 2.34 \times 10^{-11} A[/tex]

The terminal voltage at 10% of the measured current can be found by substituting I = 0.1 A in the above equation and solving for V as:

V = 0.581 V.

(b)

Given that:

V = 0.650 V

And, I = 1.00 mA.

Substituting these values in the equation above to get,

[tex]1.00 mA = Is \times (e^{0.650 V / 0.025 V} - 1)\\ Is = 4.56 \times 10^{-15} A[/tex]

The terminal voltage at 10% of the measured current can be found by substituting I = 0.1 mA in the above equation and solving for V as:

V = 0.516 V.

(c)

Given that:

V = 0.650 V

And, I = 10 μA.

Substituting these values in the equation above to get,

[tex]10 \mu A = Is \times (e^{0.650 V / 0.025 V} - 1)\\Is = 1.18 \times 10^{-16} A[/tex]

The terminal voltage at 10% of the measured current can be found by substituting I = 1 μA in the above equation and solving for V as:

V = 0.459 V.

(d)

Given that:

V = 0.700 V

And, I = 100 mA.

Substituting these values in the equation above to get,

[tex]100 \ mA = Is \times (e^{0.700 V / 0.025 V} - 1)\\Is = 2.34 \times 10^{-11} A[/tex]

The terminal voltage at 10% of the measured current can be found by substituting I = 10 mA in the above equation and solving for V as:

V = 0.581 V.

So, the values of Is and V are as: (a) [tex]Is = 2.34 \times 10^{-11} A[/tex] and V = 0.581 V. (b) [tex]Is = 4.56 \times 10^{-15} A[/tex] and V = 0.516 V. (c) [tex]Is = 1.18 \times 10^{-16} A\\[/tex] and V = 0.459 V. (d) [tex]Is = 2.34 \times 10^{-11} A[/tex] and V = 0.581 V.

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Atmospheric pressure is 101 kPa. Pressure inside a tire is measured using a typical tire pressure gage to be 900 kPa. Find gage pressure and absolute pressure in the tire. ___________________________________________________________________

Answers

Answer:

The gage and absolute pressures are 900 and 1001 kilopascals, respectively.

Explanation:

The gage pressure ([tex]P_{g}[/tex]), in kilopascals, is the difference between absolute ([tex]P_{abs}[/tex]) and atmospheric pressures ([tex]P_{atm}[/tex]), measured in kilopascals. If we know that [tex]P_{g} = 900\,kPa[/tex] and [tex]P_{atm} = 101\,kPa[/tex], then the gage and absolute pressures are, respectively:

[tex]P_{g} = 900\,kPa[/tex]

[tex]P_{abs} = P_{atm} + P_{g}[/tex]

[tex]P_{abs} = 101\,kPa + 900\,kPa[/tex]

[tex]P_{abs} = 1001\,kPa[/tex]

The gage and absolute pressures are 900 and 1001 kilopascals, respectively.

Consider the equation y = 10^(4x). Which of the following statements is true?


A plot of log(y) vs. x would be linear with a slope of 4.


A plot of log(y) vs. log (x) would be linear with a slope of 10.


A plot of log(y) vs. x would be linear with a slope of 10.


A plot of y vs. log(x) would be linear with a slope of 4.


A plot of log(y) vs. log (x) would be linear with a slope of 4.


A plot of y vs. log(x) would be linear with a slope of 10.

Answers

Answer: Plot of  [tex]\log y[/tex] vs [tex]x[/tex] would be linear with a slope  of 4.

Explanation:

Given

Equation is [tex]y=10^{4x}[/tex]

Taking log both sides

[tex]\Rightarrow \log y=4x\log (10)\\\Rightarrow \log y=4x[/tex]

It resembles with linear equation [tex]y=mx+c[/tex]

Here, slope of [tex]\log y[/tex] vs [tex]x[/tex] is 4.

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