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Tresset [83]
3 years ago
15

While discussing VIN numbers, Technician A says that the first digit of the VIN identifies the country where the vehicle was man

ufactured. Technician B says that the first digit identifies the model year of the vehicle. Who is correct
Engineering
1 answer:
ruslelena [56]3 years ago
4 0
Usually the first digit of the vin id’s the country it was built. So technician A would be correct. That’s usually how it is. Hope this helps. Please let me know if this is incorrect
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Explanation:

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What type of test can show a chemical engineer if a material remains in a system and accumulates or if it moves right through?
UkoKoshka [18]

A chemical engineer can clearly see from this kind of test if a substance stays in a system and builds up or if it just passes through.

<h3>What is a chemical engineer?</h3>
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2 years ago
Any one here play animal crossing new horizons<br> if so wanna play
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Explanation:

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3 years ago
Read 2 more answers
Create a variable pounds to store weight in pounds. Convert this to kilograms and assign the result to a variable kilos. The con
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Answer:

>>pounds=13.2

>>kilos=pounds/2.2

Explanation:

Using Matlab to write the program, consider at any time when the weight in pounds is 13.2 lb, this variable of weight is created in MATLAB by typing >>pounds=13.2. To convert it from lb to Kg, we simply divide it by 2.2 hence the second command to created is kilos. For this, the output of the program will be 6 Kg.

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A hair dryer is basically a duct in which a few layers of electric resistors are placed. A small fan pulls the air in and forces
nika2105 [10]

Answer:

a) volume flow rate of air at the inlet is 0.0471 m³/s

b) the velocity of the air at the exit is  8.517 m/s

Explanation:

Given that;

The electrical power Input W_elec = -1400 W = -1.4 kW

Inlet temperature of air T_in = 22°C

Inlet pressure of air p_in = 100 kPa

Exit temperature T_out = 47°C

Exit area of the dyer is A_out = 60 cm²= 0.006 m²

cp = 1.007 kJ/kg·K

R = 0.287 kPa·m3/kg·K

Using mass balance

m_in = m_out = m_air

W _elec = m_air ( h_in - h_out)

we know that h = CpT

so

W _elec = m_air.Cp ( T_in - T_out)

we substitute

-1.4 = m_air.1.007 ( 22 - 47 )

-1.4 =  - m_air.25.175

m_air = -1.4 / - 25.175

m_ air = 0.0556 kg/s

a) volume flow rate of air at the inlet

we know that

m_air = P_in × V_in

now from the ideal gas equation

P_in = p_in / RT_in

we substitute our values

= (100×10³) / ((0.287×10³)(22+273))

= 100000 / 84665

P_in = 1.18 kg/m³

therefore inlet volume flowrate will be;

V_in = m_air / P_in

= 0.0556 / 1.18

= 0.0471 m³/s

the volume flow rate of air at the inlet is 0.0471 m³/s

b) velocity of the air at the exit

the mass flow rate remains unchanged across the duct

m_ air = P_in.A_in.V_in = P_out.A_out.V_out

still from the ideal gas equation

P_out = p_out/ RT_out   ( assume p_in = p_out)

P_out = (100×10³) / ((0.287×10³)(47+273))

P_out  = 1.088 kg/m³

so the exit velocity will be;

V_out = m_air / P_out.A_out

we substitute our values

V_out = 0.0556 / ( 1.088 × 0.006)

= 0.0556 / 0.006528

= 8.517 m/s

 Therefore the velocity of the air at the exit is  8.517 m/s

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2 years ago
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