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mezya [45]
3 years ago
11

Technician A says that valve overlap is the number of crankshaft degrees that both valves are closed. Technician B says that val

ve overlap is the number of degrees that both valves are open. Who is​ right? A. Technician A only B. Technician B only C. Both technicians A and B D. Neither technician A nor B
Physics
1 answer:
8_murik_8 [283]3 years ago
3 0

Answer:

B. Technician B only

Explanation:

Technician A says that valve overlap is the number of crankshaft degrees that both valves are closed. Technician B says that valve overlap is the number of degrees that both valves are open. Who is​ right? A. Technician A only B. Technician B only C. Both technicians A and B D. Neither technician A nor B

Valve overlap is when  both the intake and exhaust valves are open. This occurs towards the end of the exhaust stroke, the intake valves are opened just before all the exhaust gases are released, providing more time for the intake air to enter the engine. Valve overlap must be shorter for a petrol/gasoline engine because the engine is drawing in fuel as well as air through the intake. ...it is not so for a sport car

the inlet valve and exhaust valve are located at the head of the engine. fuel enters the engine through the intake valve.

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Determine the average value of the translational kinetic energy of the molecules of an ideal gas at (a) 27.8°C and (b) 143°C. Wh
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a) k_{avg}=6.22\times 10^{-21}

b) k_{avg}=8.61\times 10^{-21}

c)  k_{mol}=3.74\times 10^{3}J/mol

d)   k_{mol}=5.1\times 10^{3}J/mol

Explanation:

Average translation kinetic energy (k_{avg}) is given as

k_{avg}=\frac{3}{2}\times kT    ....................(1)

where,

k = Boltzmann's constant ; 1.38 × 10⁻²³ J/K

T = Temperature in kelvin

a) at T = 27.8° C

or

T = 27.8 + 273 = 300.8 K

substituting the value of temperature in the equation (1)

we have

k_{avg}=\frac{3}{2}\times 1.38\times 10^{-23}\times 300.8  

k_{avg}=6.22\times 10^{-21}J

b) at T = 143° C

or

T = 143 + 273 = 416 K

substituting the value of temperature in the equation (1)

we have

k_{avg}=\frac{3}{2}\times 1.38\times 10^{-23}\times 416  

k_{avg}=8.61\times 10^{-21}J

c ) The translational kinetic energy per mole of an ideal gas is given as:

       k_{mol}=A_{v}\times k_{avg}

here   A_{v} = Avagadro's number; ( 6.02×10²³ )

now at T = 27.8° C

        k_{mol}=6.02\times 10^{23}\times 6.22\times 10^{-21}

          k_{mol}=3.74\times 10^{3}J/mol

d) now at T = 143° C

        k_{mol}=6.02\times 10^{23}\times 8.61\times 10^{-21}

          k_{mol}=5.1\times 10^{3}J/mol

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