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mina [271]
3 years ago
11

I need help on this

Engineering
1 answer:
Over [174]3 years ago
3 0

Answer:

b.) The car travels around a circular track at 30 m/s

Explanation:

To find - In which situation is the speed of the car constant while its velocity is changing?

a.) The car travels down a straight track at 30 m/sec

b.) The car travels around a circular track at 30 m/s

c.) The car begins from rest and accelerates to 20 m/s

d.) The car begins traveling at 30 m/s and decreases to 15 m/s

Proof -

We know that,

Speed is distance covered in a particular time

Velocity is distance covered by body in particular time in defined direction.

Now,

In a.)

Speed and direction is not changing.

So, Option a) is wrong.

In c.) and d.)

There exists a  positive and negative acceleration.

Even If direction is constant, but the magnitude of speed is changing here, So, Option c) and d) are wrong.

Now,

In b.)

We can see that,

Due to circular motion, the speed of the car is constant at 30 m/s

But Due to circular motion, direction is changing continuously.

So, here speed is constant while its velocity is changing.

∴

The correct option is - b.) The car travels around a circular track at 30 m/s

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ear shaft.3. Chapter 12 –Loading on Spur Gears: A 26-tooth pinion rotating at a uniform 1800 rpm meshes with a 55-tooth gear in
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Answer:

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Explanation:

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V_{t} =\frac{0.13*2*\pi *1800}{120} =12.25m/s

The factor B is equal to:

B=\frac{(12-Q_{v})^{2/3}  }{4} , if Q_{v} =10\\B=\frac{(12-10)^{2/3} }{4} =0.396

The factor A is equal to:

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The dynamic factor is:

K_{v} =(\frac{A}{A+\sqrt{200V_{t} } } )^{B} \\K_{v}=(\frac{83.82}{83.82+\sqrt{200*12.25} } )^{0.396} =0.832

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JR = 0.41

Ka = 1

Kb = 1

Ks = 1

Ki = 1.42

Km = 1.7

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P_{d} =\frac{1}{m} =\frac{1}{5} =0.2mm^{-1}

The bending stress is equal to:

\sigma =\frac{W_{t}P_{d}K_{a}K_{m}K_{s}K_{b}K_{i} }{FJ_{g}K_{v}}  \\\sigma =\frac{22000*0.2*1*1.7*1*1*1.42}{62*0.41*0.832} =502.22MPa

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Answer:

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Explanation:

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