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AnnZ [28]
3 years ago
9

In which one of the following four-stroke cylinder arrangements will the intake and exhaust valves open downward into the combus

tion chamber?
A. K-type
B. C-type
C. T-type
D. I-type
Physics
1 answer:
Bogdan [553]3 years ago
8 0
The  cylinder arrangements that will do the job is: <span>D. I-type 

In the i-type cylinder arrangement, </span><span> the intake and exhaust valves  are positioned vertically at the top of the cylinder head, and both valves will move downward for the combustion chamber to open.
We could see this type of cylinder arrangement on many automobiles product</span>
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A railroad car of mass 2.50*10^4 kg is moving with a speed of 4.00 m/s. It collides and couples with three other coupled railroa
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Answer:

(a) 2.5 m/s

(b) 37.5 KJ

Explanation:

(a)

From the law of conservation of momentum, Initial momentum=Final momentum

mV_1+3mV_2=(m+3m)V_f=4mV_f

V_1+3V_2=4V_f and making V_f the subject then

V_f=\frac {V_1+3V_2}{4} and since V_1 is initial velocity of car, value given as 4 m/s, V_2 is the initial velocity of the three cars stuck together, value given as 2 m/s and v_f is the final velocity which is unknown. By substitution

V_f=\frac {4+(3\times2)}{4}=2.5 m/s

(b)

Initial kinetic energy is given by

\frac {mV_1^{2}}{2}+\frac {3mV_2^{2}}{2}=\frac {m(V_1^{2}+3V_2^{2}}{2}=\frac {2.5\times 10^{4}(4^{2}+3(2^{2}))}{2}=350\times10^{3} J= 350 KJ

Final kinetic energy is given by

\frac {4mV_f^{2}}{2}=\frac {4\times 2.5\times 10^{4}\times 2.5^{2}}{2}=312.5\times 10^{3} J=312.5 KJ

The energy lost is given by subtracting the final kinetic energy from the initial kinetic energy hence

Energy lost=350-312.5=37.5 KJ

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Answer:

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KE=0.5*50*2^{2}=100 J

Therefore, the child's kinetic energy is equivalent to 100 J

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Answer is C

Explanation:

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