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yaroslaw [1]
3 years ago
14

Which of the following can cause a flopping sound at the front of the engine

Engineering
1 answer:
Irina-Kira [14]3 years ago
3 0
B) timing chain too loose
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DiKsa [7]

Would it be Unit?

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5 0
3 years ago
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The drag coefficient of a vehicle increases when its windows are rolled down or its sunroof is opened. A sports car has a fronta
klio [65]

Answer:

Explanation:

The additional power consumption of the car when v :

35 mi/h = 0.464hp

70 mi/h = 3.71hp.

For the step by step explanation of how we arrived at these answers, please go through the attached files.

3 0
3 years ago
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Indicates the design of the building and<br> adjoining areas
Lelu [443]

Answer:

Architectural Plan

Explanation:

4 0
3 years ago
At an axial load of 22 kN, a 15-mm-thick × 40-mm-wide polyimide polymer bar elongates 4.1 mm while the bar width contracts 0.15
Alenkasestr [34]

Answer:

The Poisson's Ratio of the bar is 0.247

Explanation:

The Poisson's ratio is got by using the formula

Lateral strain / longitudinal strain

Lateral strain = elongation / original width (since we are given the change in width as a result of compession)

Lateral strain = 0.15mm / 40 mm =0.00375

Please note that strain is a dimensionless quantity, hence it has no unit.

The Longitudinal strain is the ratio of the elongation to the original length in the longitudinal direction.

Longitudinal strain = 4.1 mm / 270 mm = 0.015185

Hence, the Poisson's ratio of the bar is 0.00375/0.015185 = 0.247

The Poisson's Ratio of the bar is 0.247

Please note also that this quantity also does not have a dimension

3 0
3 years ago
An FPC 4 m2 in area is tested during the night to measure the overall heat loss coefficient. Water at 60 C circulates through th
sp2606 [1]

Answer:

<em> - 14.943 W/m^2K  ( negative sign indicates cooling ) </em>

Explanation:

Given data:

Area of FPC = 4 m^2

temp of water = 60°C

flow rate = 0.06 l/s

ambient temperature = 8°C

exit temperature = 49°C

<u>Calculate the overall heat loss coefficient </u>

Note : heat lost by water = heat loss through convection

m*Cp*dT  = h*A * ( T - To )

∴ dT / T - To = h*A / m*Cp  ( integrate the relation )

In ( \frac{49-8}{60-8} ) =  h* 4 / ( 0.06 * 10^-3 * 1000 * 4180 )

In ( 41 / 52 ) = 0.0159*h

hence h = - 0.2376 / 0.0159

              = - 14.943  W/m^2K  ( heat loss coefficient )

7 0
3 years ago
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