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Rainbow [258]
3 years ago
6

Which option demonstrates when most vehicles lose their efficiency?

Engineering
2 answers:
faltersainse [42]3 years ago
7 0

Answer:

During starts and stops

Explanation:

"Many of the energy inefficiencies in vehicles occur during starts and stops."

found it in the text.

guapka [62]3 years ago
6 0

Hey there ..

I think the answer is as they put brake ..

I am not sure .. just check the image also provided above .. ..

If u think this helped u ..plz mark me as brainliest ..

And follow me

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A motor car shaft consists of a steel tube 30 mm internal diameter and 4 mm thick. The engine develops 10 kW at 2000 r.p.m. Find
tresset_1 [31]

The maximum shear stress in the tube when the power is transmitted through a 4: 1 gearing is 28.98 MPa.

<h3>What is power?</h3>

Power is the energy transferred per unit time.

Torque is find out by

P = 2πNT/60

10000 = 2π x 2000 x T / 60

T =47.74 N.m

The gear ratio Ne / Ns =4/1

Ns =2000/4 = 500

Ts =Ps x 60/(2π x 500)

Ts =190.96 N.m

Maximum shear stress τ = 16/π x (T / (d₀⁴ - d₁⁴))

τ max =T/J x D/2
where d₁ = 30mm = 0.03 m

           d₀ = 30 +(2x 4) = 38mm =0.038 m

Substitute the values into the equation, we get

τ max = 16 x 190.96 x 0.038 /π x (0.038⁴ - 0.03⁴)

τ max = 28.98 MPa.

Thus, the maximum shear stress in the tube is 28.98 MPa.

Learn more about power.

brainly.com/question/13385520

#SPJ1

7 0
2 years ago
True or false. Part of the mission of the NTSB is to determine the probable cause of an accident
eimsori [14]

uniform

welcome 2 Ghana African state western region

6 0
3 years ago
Best use for a suspension bridge
babunello [35]

Answer:

over a rive or fast moving water or canyon

Explanation: you would use a suspension bridge in an area where you can't put supports down.

4 0
3 years ago
Water is flowing at a rate of 0.15 ft3/s in a 6 inch diameter pipe. The water then goes through a sudden contraction to a 2 inch
Georgia [21]

Answer:

Head loss=0.00366 ft

Explanation:

Given :Water flow rate Q=0.15 \frac{ft^{3}}{sec}

         D_{1}= 6 inch=0.5 ft

        D_{2}=2 inch=0.1667 ft

As we know that Q=AV

A_{1}\times V_{1}=A_{2}\times V_{2}

So V_{2}=\frac{Q}{A_2}

     V_{2}=\dfrac{.015}{\frac{3.14}{4}\times 0.1667^{2}}

     V_{2=0.687 ft/sec

We know that Head loss due to sudden contraction

           h_{l}=K\frac{V_{2}^2}{2g}

If nothing is given then take K=0.5

So head lossh_{l}=(0.5)\frac{{0.687}^2}{2\times 32.18}

                                    =0.00366 ft

So head loss=0.00366 ft

4 0
3 years ago
The angle formed between the SAI and the camber line is called the
hichkok12 [17]

Answer:C

Explanation:

8 0
3 years ago
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