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3 years ago

An organization sets its standards for quality according to the best product it can produce.

Engineering

2 answers:

Marianna [84]3 years ago

6 0

I believe it’s True, but please correct me if I’m wrong!

dusya [7]3 years ago

5 0

Answer:

True!! took the test

Explanation:

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How is the fuel introduced into the Diesel engine?

Ugo [173]

Answer:

diesel fuel is pumped at high pressure to the injectors which are responsible for entering the fuel into the combustion chamber,

when the piston is at the top the pressure is so high that it explodes the fuel (diesel) that results in a generation of mechanical power

5 0

3 years ago

What is the area enclosed by the cycle area of the Carnot cycle illustrating on a P-V diagram?

Inga [223]

Answer:

The work of the cycle.

Explanation:

The area enclosed by the cycle of the Pressure-Volume diagram of a Carnot engine represents the net work performed by the cycle.

The expansions yield work, and this is represented by the area under the curve all the way to the p=0 line. But the compressions consume work (or add negative work) and this is substracted fro the total work. Therefore the areas under the compressions are eliminated and you are left with only the enclosed area.

7 0

3 years ago

Assume a program requires the execution of 50 x 106 FP instructions, 110 x 106 INT instructions, 80 x 106 L/S instructions, and

Pavlova-9 [17]

Answer:

Part A:

$1.3568*10^{-5}=\frac{5300* New\ CPI_1+11660*1+8480*4+1696*2}{2*10^9\ Hz} \\ New\ CPI_1=-4.12$

CPI cannot be negative so it is not possible to for program to run two times faster.

Part B:

$1.3568*10^{-5}=\frac{5300*1+11660*1+8480*New\ CPI_3+1696*2}{2*10^9\ Hz} \\ New\ CPI_3=0.8$

CPI reduced by $1-\frac{0.8}{4} = 0.80$ =80%

Part C:

New Execution Time= $\frac{5300*0.6+11660*0.6+8480*2.8+1696*1.4}{2*10^9\ Hz}=1.81472*10^{-5}\ s$

Increase in speed= $1-\frac{1.81472*10^{-5}}{2.7136*10^{-5}} =0.33125= 33.125\%$

Explanation:

FP Instructions=50*106=5300

INT Instructions=110*106=11660

L/S Instructions=80*106=8480

Branch Instructions=16*106=1696

Calculating Execution Time:

Execution Time= $\frac{\sum^4_{i=1} Number\ of\ Instruction*\ CPI_{i}}{Clock\ Rate}$

Execution Time= $\frac{5300*1+11660*1+8480*4+1696*2}{2*10^9\ Hz}$

Execution Time= $2.7136*10^{-5}\ s$

Part A:

For Program to run two times faster,Execution Time (Calculated above) is reduced to half.

New Execution Time= $\frac{2.7136*10^{-5}}{2}=1.3568*10^{-5}\ s$

$1.3568*10^{-5}=\frac{5300* New\ CPI_1+11660*1+8480*4+1696*2}{2*10^9\ Hz} \\ New\ CPI_1=-4.12$

CPI cannot be negative so it is not possible to for program to run two times faster.

Part B:

For Program to run two times faster,Execution Time (Calculated above) is reduced to half.

New Execution Time= $\frac{2.7136*10^{-5}}{2}=1.3568*10^{-5}\ s$

$1.3568*10^{-5}=\frac{5300*1+11660*1+8480*New\ CPI_3+1696*2}{2*10^9\ Hz} \\ New\ CPI_3=0.8$

CPI reduced by $1-\frac{0.8}{4} = 0.80$ =80%

Part C:

$New\ CPI_1=0.6*Old\ CPI_1=0.6*1=0.6\\New\ CPI_2=0.6*Old\ CPI_2=0.6*1=0.6\\New\ CPI_3=0.7*Old\ CPI_3=0.7*4=2.8\\New\ CPI_4=0.7*Old\ CPI_4=0.7*2=1.4$

New Execution Time= $\frac{\sum^4_{i=1} Number\ of\ Instruction*\ CPI_{i}}{Clock\ Rate}$

New Execution Time= $\frac{5300*0.6+11660*0.6+8480*2.8+1696*1.4}{2*10^9\ Hz}=1.81472*10^{-5}\ s$

Increase in speed= $1-\frac{1.81472*10^{-5}}{2.7136*10^{-5}} =0.33125= 33.125\%$

8 0

3 years ago

rt this question on a NEW page Give ONE word/tem for each of the following descriptions by choosing a word/term from the list be

Shalnov [3]

Answer:

oh I am sorry I can't understand your question.

6 0

3 years ago

what would my engine exhaust sound like if i made a custom muffler that was just smooth on the inside, would it have an echoing

Elena L [17]

Answer:

you know what sounds rlly nice? first have you seent he new corvettes? next u should here a nisan gtr. (like tanner foxes)

Explanation:

8 0

3 years ago