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vovikov84 [41]
2 years ago
14

Describe how the Rotary Engine works.

Engineering
1 answer:
LekaFEV [45]2 years ago
8 0

Answer:

  Rotary engine was early known by the name of internal combustion engine. It convert heat from a high pressure of combustion. The main advantage of rotary engine is that it can be operate with less number of vibration. It works on the principle of converting pressure into rotating motion. In rotary engine the expansion pressure is applied on the flank rotor.  

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HW6P2 (20 points) The recorded daily temperature (°F) in New York City and in Denver, Colorado during the month of January 2014
Maurinko [17]

Answer & Explanation:

function Temprature

NYC=[33 33 18 29 40 55 19 22 32 37 58 54 51 52 45 41 45 39 36 45 33 18 19 19 28 34 44 21 23 30 39];

DEN=[39 48 61 39 14 37 43 38 46 39 55 46 46 39 54 45 52 52 62 45 62 40 25 57 60 57 20 32 50 48 28];

%AVERAGE CALCULATION AND ROUND TO NEAREST INT

avgNYC=round(mean(NYC));

avgDEN=round(mean(DEN));

fprintf('\nThe average temperature for the month of January in New York city is %g (F)',avgNYC);

fprintf('\nThe average temperature for the month of January in Denvar is %g (F)',avgDEN);

%part B

count=1;

NNYC=0;

NDEN=0;

while count<=length(NYC)

   if NYC(count)>avgNYC

       NNYC=NNYC+1;

   end

   if DEN(count)>avgDEN

        NDEN=NDEN+1;

   end

   count=count+1;

end

fprintf('\nDuring %g days, the temprature in New York city was above the average',NNYC);

fprintf('\nDuring %g days, the temprature in Denvar was above the average',NDEN);

%part C

count=1;

highDen=0;

while count<=length(NYC)

   if NYC(count)>DEN(count)

       highDen=highDen+1;

   end

   count=count+1;

end

fprintf('\nDuring %g days, the temprature in Denver was higher than the temprature in New York city.\n',highDen);

end

%output

check the attachment for additional Information

8 0
3 years ago
What is the IMA of this pulley belt system if the diameter of the input
Stella [2.4K]

Answer:

2.8

Explanation:

The ideal mechanical advantage of the pulley IMA  = D'/D where D' = diameter of output pulley = 7 inches and D = diameter of input pulley = 2.5 inches

So, IMA = D'/D

= 7/2.5

= 2.8

So, the ideal mechanical advantage of the pulley IMA = 2.8

8 0
2 years ago
before adjusting drive-belt tension, technician a checks for proper pulley alignment. technician b looks up the specified belt t
Vsevolod [243]

Answer:

Technician b is correct

Explanation:

Before adjusting drive-belt tension, it is very important to check the vehicle workshop manual for specified belt tension, so that you can match your reading against the specification in the vehicle's service manual. If the tension reading you have matches the suggested reading in the vehicle's service manual and the belt is not damaged then you do not need to proceed any further. But if the reading does not match, then you can adjust the belt tension.

Therefore, technician b is correct.

5 0
3 years ago
Determine the maximum mass of the crate so
Tpy6a [65]

Answer:

293 kg

Explanation:

Let's say the tension in each cable is Tb, Tc, and Td.

First, find the length of cable AD:

r = √(2² + 2² + 1²)

r = 3

Using similar triangles:

Tdx = 2/3 Td

Tdy = 2/3 Td

Tdz = 1/3 Td

Sum of the forces in the x direction:

∑F = ma

Tb − 2/3 Td = 0

Td = 3/2 Tb

Sum of the forces in the y direction:

∑F = ma

2/3 Td − Tc = 0

Td = 3/2 Tc

Sum of the forces in the z direction:

∑F = ma

1/3 Td − mg = 0

Td = 3mg

From the first two equations, we know Td is greater than Tb or Tc.  So we need to set Td to 8.6 kN, or 8600 N.

8600 N = 3mg

m = 8600 N / (3 × 9.8 m/s²)

m ≈ 292.5 kg

Rounded to three significant figures, the maximum mass of the crate is 293 kg.

7 0
3 years ago
A strip ofmetal is originally 1.2m long. Itis stretched in three steps: first to a length of 1.6m, then to 2.2 m, and finally to
andreyandreev [35.5K]

Answer:

strains for the respective cases are

0.287

0.318

0.127

and for the entire process 0.733

Explanation:

The formula for the true strain is given as:

\epsilon =\ln \frac{l}{l_{o}}

Where

\epsilon = True strain

l= length of the member after deformation

l_{o} = original length of the member

<u>Now for the first case we have</u>

l= 1.6m

l_{o} = 1.2m

thus,

\epsilon =\ln \frac{1.6}{1.2}

\epsilon =0.287

<u>similarly for the second case we have</u>

l= 2.2m

l_{o} = 1.6m   (as the length is changing from 1.6m in this case)

thus,

\epsilon =\ln \frac{2.2}{1.6}

\epsilon =0.318

<u>Now for the third case</u>

l= 2.5m

l_{o} = 2.2m

thus,

\epsilon =\ln \frac{2.5}{2.2}

\epsilon =0.127

<u>Now the true strain for the entire process</u>

l=2.5m

l_{o} = 1.2m

thus,

\epsilon =\ln \frac{2.5}{1.2}

\epsilon =0.733

6 0
2 years ago
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