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uysha [10]
3 years ago
8

Driving test. Need NOW

Engineering
1 answer:
Flauer [41]3 years ago
3 0

Answer:

4 - Double Parking

6- Perpendicular Parking

5- Hand over hand method

1 - Manual Transmission

3 - Push pull method

2 - Parking on a hill

Explanation:

Manual Transmission - requires a stick shift

Perpendicular Parking - type of parking in a parking lot.

Double Parking - Parking in the roadway

Hand-over-hand - Grab the far side of the wheel.

Push-pull Method - Slide your hand along the wheel.

Parking on a hill - Turn wheels toward curb.

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Air flows through a 0.25-m-diameter duct. At the inlet the velocity is 300 m/s, and the stagnation temperature is 90°C. If the M
Naddika [18.5K]

Answer:

a. 318.2k

b. 45.2kj

Explanation:

Heat transfer rate to an object is equal to the thermal conductivity of the material the object is made from, multiplied by the surface area in contact, multiplied by the difference in temperature between the two objects, divided by the thickness of the material.

See attachment for detailed analysis

7 0
4 years ago
Showing all of your work and algebra,generate an approximate expression for T as a function ofthe other variables. (b) Explain w
shusha [124]

Answer:

Following the ways of dealing with incomplete questions, i was able to get the complete question, please look at the attachment for ans.

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4 years ago
Exhaust gas from a furnace is used to preheat the combustion air supplied to the furnace burners. The gas, which has a flow rate
Monica [59]

Answer:

The total tube surface area in m² required to achieve an air outlet temperature of 850 K is 192.3 m²

Explanation:

Here we have the heat Q given as follows;

Q = 15 × 1075 × (1100 - t_{A2}) = 10 × 1075 × (850 - 300) = 5912500 J

∴ 1100 - t_{A2} = 1100/3

t_{A2}  = 733.33 K

\Delta \bar{t}_{a} =\frac{t_{A_{1}}+t_{A_{2}}}{2} - \frac{t_{B_{1}}+t_{B_{2}}}{2}

Where

\Delta \bar{t}_{a} = Arithmetic mean temperature difference

t_{A_{1} = Inlet temperature of the gas = 1100 K

t_{A_{2} = Outlet temperature of the gas = 733.33 K

t_{B_{1} =  Inlet temperature of the air = 300 K

t_{B_{2} = Outlet temperature of the air = 850 K

Hence, plugging in the values, we have;

\Delta \bar{t}_{a} =\frac{1100+733.33}{2} - \frac{300+850}{2} = 341\tfrac{2}{3} \, K = 341.67 \, K

Hence, from;

\dot{Q} = UA\Delta \bar{t}_{a}, we have

5912500  = 90 × A × 341.67

A = \frac{5912500  }{90 \times 341.67} = 192.3 \, m^2

Hence, the total tube surface area in m² required to achieve an air outlet temperature of 850 K = 192.3 m².

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3 years ago
Which of the following is a variable expense for many adults?
sertanlavr [38]
But where are the options?
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3 years ago
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Safety measures to be taken during technical drawing<br>​
Pavlova-9 [17]

Explanation:

Couldn't you just leave it in centimeters?

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