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3 years ago

If a steel cable is rated to take 800-lb and the steel has a yield strength of 90,000psi, what is the diameter of the cable?

Engineering

1 answer:

goldfiish [28.3K]3 years ago

5 0

Answer:

d = 2.69 mm

Explanation:

Assuming the cable is rated with a factor of safety of 1.

The stress on the cable is:

σ = P/A

Where

σ = normal stress

P: load

A: cross section

The section area of a circle is:

A = π/4 * d^2

Then:

σ = 4*P / (π*d^2)

Rearranging:

d^2 = 4*P / (π*σ)

$d = \sqrt{4*P / (\pi*\sigma)}$

Replacing:

$d = \sqrt{4*800 / (\pi*\90000)} = 0.106 inches$

0.106 inches = 2.69 mm

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Two technicians are discussing the intake air temperature (IAT) sensor. Technician A says that the computer uses the IAT sensor

mart [117]

Both the technicians are correct.

Explanation

Intake air temperature sensor is used in engines of vehicles to monitor the temperature of air entering the engine.

They are basically made of thermistors whose electrical resistance changes according to temperature.

Depending upon the reading and accuracy of intake air temperature sensor, the power-train control module (PCM) will decide about the air and fuel mixture ratio in the engine.

The hot air in engine requires less fuel to operate the engine parts while cold air requires more fuel to operate the engine.

The ratio of air and fuel mixture should be maintained in the engine and it is done by PCM only after getting the input from IAT. So technician B is saying correct.

Also the IAT works as a backup to support the engine coolant temperature sensor by the computer.

As the IAT checks the temperature of outside air, it will help to change the coolant temperature of the engine based on the environment.

Thus technician A is also correct. So both the technicians are correct.

6 0

3 years ago

Air at a pressure of 6000 N/m^2 and a temperature of 300C flows with a velocity of 10 m/sec over a flat plate of length 0.5 m. E

White raven [17]

Answer:

$Q=hA(T_{w}-T_{inf})=16.97*0.5(27-300)=-2316.4J$

Explanation:

To solve this problem we use the expression for the temperature film

$T_{f}=\frac{T_{\inf}+T_{w}}{2}=\frac{300+27}{2}=163.5$

Then, we have to compute the Reynolds number

$Re=\frac{uL}{v}=\frac{10\frac{m}{s}*0.5m}{16.96*10^{-6}\rfac{m^{2}}{s}}=2.94*10^{5}$

Re<5*10^{5}, hence, this case if about a laminar flow.

Then, we compute the Nusselt number

$Nu_{x}=0.332(Re)^{\frac{1}{2}}(Pr)^{\frac{1}{3}}=0.332(2.94*10^{5})^{\frac{1}{2}}(0.699)^{\frac{1}{3}}=159.77$

but we also now that

$Nu_{x}=\frac{h_{x}L}{k}\\h_{x}=\frac{Nu_{x}k}{L}=\frac{159.77*26.56*10^{-3}}{0.5}=8.48\\$

but the average heat transfer coefficient is h=2hx

h=2(8.48)=16.97W/m^{2}K

Finally we have that the heat transfer is

$Q=hA(T_{w}-T_{inf})=16.97*0.5(27-300)=-2316.4J$

In this solution we took values for water properties of

v=16.96*10^{-6}m^{2}s

Pr=0.699

k=26.56*10^{-3}W/mK

A=1*0.5m^{2}

I hope this is useful for you

regards

8 0

3 years ago

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kozerog [31]

Answer:

Explanation:Come

tomate

5 0

2 years ago

Read 2 more answers

Expalin the application of diesel cycle in detail.

mars1129 [50]

Explanation:

Diesel cycle:

All diesel engine work on diesel cycle .In diesel cycle there are four process .These processes are as follows

1. Adiabatic reversible compression

2.Heat addition at constant pressure

3.Adiabatic reversible expansion

4.Constant volume heat rejection

In general compression ratio in diesel engine is high as compare to petrol engine.But the efficiency of diesel cycle is less as compare to petrol cycle for same compression ratio.

Applications of diesel cycle:

Generally diesel cycle used for heavy vehicle or equipment because heavy vehicle or equipment is required high initial torque.So this cycle have lots of applications such as in industrial machining,in trucks,power plant,in mining ,in defense or military,large motors ,compressor and pump etc.

5 0

3 years ago

1. An automobile travels along a straight road at 15.65 m/s through a 11.18 m/s

djyliett [7]

Answer:

a.) Time = 17.13 seconds

b.) 31.88 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

8 0

3 years ago