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Serggg [28]
4 years ago
14

Number the statements listed below in the order that they would occur in engine operation. Then, label these stages as intake, c

ompression, power, or exhaust.
a. The fuel mixture is ignited by the spark plug, and the resulting explosion forces the piston down in the cylinder.

b. The piston rises and compresses the fuel mixture in the combustion chamber.

c. The exhaust gases left over from the combustion are released from the cylinder. d. A mixture of air and fuel is drawn into the cylinder
Engineering
1 answer:
boyakko [2]4 years ago
6 0

Answer:

d. A mixture of air and fuel is drawn into the cylinder ( Intake)

b. The piston rises and compresses the fuel mixture in the combustion chamber.(compression)

a. The fuel mixture is ignited by the spark plug, and the resulting explosion forces the piston down in the cylinder.(power)

c. The exhaust gases left over from the combustion are released from the cylinder.(exhaust)

Explanation:

The first step in engine operation is called <u>the intake stage</u>, where air mixed with fuel is drawn into the cylinder. This is followed by the <u>compression stage </u>where the piston rises to compress the air mixed with fuel trapped in the chamber.<u>The power stage</u> follows, the air-fuel mixture ignites to cause a spark and small explosion enough to push the piston back down cylinder causing a motion to develop down the piston to the crankshaft.<u>The exhaust stag</u>e is the final one which sees the exhaust gases  released from the cylinder.

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An automobile having a mass of 1100 kg initially moves along a level highway at 120 km/h relative to the highway. It then climbs
soldier1979 [14.2K]

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-6111.11\ \text{kJ}

863.28\ \text{kJ}

Explanation:

m = Mass of automobile = 1100 kg

v = Velocity of car = 120 km/h = \dfrac{120}{3.6}\ \text{m/s}

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Change in kinetic energy

KE=\dfrac{1}{2}m(u^2-v^2)\\\Rightarrow KE=\dfrac{1}{2}\times 1100\times (0-(\dfrac{120}{3.6})^2)\\\Rightarrow KE=-611111.11\ \text{J}

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Change in potential energy is given by

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8 0
3 years ago
Twenty distinct cars park in the same parking lot every day. Ten of these cars are US-made, while the other ten are foreign-made
Zina [86]

Answer:

Total no. of ways to line up cars is 20! = 2.43 c 10^18

Probability that the cars alternate is 0.00001 or 0.001%

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For the probability that cars alternate, two groups will be formed, one consisting of US-made 10 cars and other containing 10 foreign made. The number of favorable outcomes for this can be found out as the arrangements of 2! between these groups multiplied by the arrangements of 10! for each group, due to the arrangements among the groups themselves.

Favorable Outcomes = 2! x 10! x 10!

Thus the probability of event will be:

Probability = Favorable Outcomes/Total No. of Ways

Probability = (2! x 10! x 10!)/20!

<u>Probability = 0.00001 = 0.001%</u>

4 0
3 years ago
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