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3 years ago

What type of lights should you use if a vehicle is approaching you from the 1 point

Engineering

1 answer:

Pepsi [2]3 years ago

5 0

Answer:

dim lights

Explanation:

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Proper handling of blueprints includes which of the following

marta [7]

Answer:

folding plans neatly after use

3 0

3 years ago

Impact strips may be designed into a bumper cover.<br> True<br> False

d1i1m1o1n [39]

Answer:

true I think

Explanation:

true I think

5 0

3 years ago

Saturated water vapor undergoes a throttling process from 1bar to a 0.35bar. What is the change in temperature for this process?

mamaluj [8]

Answer:

-25.63°C.

Explanation:

We know that throttling is a constant enthalpy process

$h_1=h_2$

From steal table

We know that if we know only one property in side the dome then we will find the other property by using steam property table.

Temperature at saturation pressure 1 bar is 99.63°C and Temperature at saturation pressure 0.35 bar is about 74°C .

So from above we can say that change in temperature is -25.63°C.

But there is no any option for that .

4 0

3 years ago

A gasoline engine has a piston/cylinder with 0.1 kg air at 4 MPa, 1527◦C after combustion, and this is expanded in a polytropic

Roman55 [17]

Answer:

The expansion work is 71.24 kJ and heat transfer is -16.89 kJ

Explanation:

From ideal gas law,

Initial volume (V1) = nRT/P

n is the number of moles of air in the cylinder = mass/MW = 0.1/29 = 0.00345 kgmol

R is gas constant = 8314.34 J/kgmol.K

T is initial temperature = 1527 °C = 1527+273 = 1800 K

P is initial pressure = 4 MPa = 4×10^6 Pa

V1 = 0.00345×8314.34×1800/(4×10^6) = 0.013 m^3

V2 = 10×V1 = 10×0.013 = 0.13 m^3

The process is a polytropic expansion process

polytropic exponent (n) = 1.5

P2 = P1(V1/V2)^n = 4×10^6(0.013/0.13)^1.5 = 1.26×10^5 Pa

Expansion work = (P1V1 - P2V2) ÷ (n - 1) = (4×10^6 × 0.013 - 1.26×10^5 × 0.13) ÷ (1.5 - 1) = 35620 ÷ 0.5 = 71240 J = 71240/1000 = 71.24 kJ

Heat transfer = change in internal energy + expansion work

change in internal energy (∆U) = Cv(T2 - T1)

T2 = PV/nR = 1.26×10^5 × 0.13/0.00345×8314.34 = 571 K

Cv = 20.785 kJ/kgmol.K

∆U = 20.785(571 - 1800) = -25544.765 kJ/kgmol × 0.00345 kgmol = -88.13 kJ

Heat transfer = -88.13 + 71.24 = -16.89 kJ

5 0

3 years ago

Marks

Lina20 [59]

Answer:

i think C

Explanation:

if not really sorry

3 0

2 years ago