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astra-53 [7]
3 years ago
7

6. When the engine stalls or the power unit fails, on a car with power

Engineering
1 answer:
frutty [35]3 years ago
3 0
B

But

I think

So yea it prob isn’t
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Hey answr this sajida Yusof
Allisa [31]
What do u need? Rusbaisuwvwbs
5 0
3 years ago
Read 2 more answers
If the dry-bulb temperature is 95°F and the wet-bulb temperature is also 78°F, what is the relative humid- ity? What is the dew
Sidana [21]

Answer:

Relative humidity 48%.

Dew point 74°F

humidity ratio 118 g of moisture/pound of dry air

enthalpy 41,8 BTU per pound of dry air

Explanation:

You can get this information from a Psychrometric chart for water, like the one attached.

You enter the chart with dry-bulb and wet-bulb temperatures (red point in the attachment) and following the relative humidity curves you get approximately 48%.

To get the dew point you need to follow the horizontal lines to the left scale (marked with blue): 74°F

for the humidity ratio you need to follow the horizontal lines but to the rigth scale (marked with green): 118 g of moisture/pound of dry air

For enthalpy follow the diagonal lines to the far left scale (marked with yellow): 41,8 BTU per pound of dry air

5 0
3 years ago
How many gallons of water can you collect on a roof 40'×35' in a 1" rain?​
Vedmedyk [2.9K]

The answer to your question should be 630, if you have a roof the size of 40 feet, by 35 feet.

If the roof is 40 inches, by 35 inches, you would have collected 4.36 inches of rain.

Explanation: To calculate this you are going to do this:

Length of the roof ______ feet * width of roof _______ feet * .6 gallons per square feet *.75 * ________ however many inches of rain

* means multiply

7 0
3 years ago
Why is it important for engineers to consider both short and long term implications of their work?
polet [3.4K]
The right answer should be A).
7 0
3 years ago
The minimum fresh air requirement of a residential building is specified to be 0.35 air changes per hour (ASHRAE, Standard 62, 1
Thepotemich [5.8K]

Answer:

a) The flow capacity of the fan is 3150 L/min

d) the minimum diameter is 0.11 m

Explanation:

given data:

A = area of residence = 200 m²

h = height of building = 2.7 m

Percentage of air that must be replaced by fresh air is 35%

v = velocity of air in the duct = 5.5 m/s

a) The volume of the entire building is:

Volume=2.7*200=540m^{3}

The flow capacity of the fan is equal to:

Flow=\frac{0.35*540}{60} =3.15m^{3} /min

Flow=3.15\frac{m^{3} }{min} *\frac{1000L}{1m^{3}  } =3150L/min

b) The volume flow rate of fresh air is equal to:

Flow=\frac{\pi *d^{2} }{4} V\\d=\sqrt{\frac{4*Flow}{\pi V} } =\sqrt{\frac{4*3.15}{\pi *5.5*60} } =0.11m

6 0
4 years ago
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