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8090 [49]
2 years ago
10

The pressure gauge on a tank of CO2 used to fill Dora-water bottles reads 51.0 psi.

Engineering
1 answer:
tensa zangetsu [6.8K]2 years ago
3 0

Known :

Pg = 51 psi

Patm = 28 inHg = 13.7 psi

Solution :

Absolute pressure is gauge pressure plus the atmospheric pressure :

Pabs = Pg + Patm

Pabs = 51 + 13.7

Pabs = 64.8 psi

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John has an exhaust leak in his Acura Integra GS-R, What steps would he take to fix the leak in time for his inspection?
SCORPION-xisa [38]

Answer:

Explanation:

Fist you need to identify where the leak is coming from. You can do this by either listening for the leak or spraying soapy water on the exhaust to look for air bubbles coming out of the exhaust. Depending on the spot of the leak there are many ways you can fix this leak.

1. Exhaust clamp

2. Exhaust putty

3. Exhaust tape

4. New exhaust

Exhaust clamp is best used for holes on straight pipes.

Putty is best used on welds or small holes like on exhaust manifolds or welds connecting various pieces like catalytic converters, mufflers, or resonators.

Tape will work best on straight pipes with holes.

New exhaust is for when the thig is beyond repair, like rust.

Now good luck because working on exhausts is a pain.

4 0
2 years ago
A pitfall cited in Section 1.10 is expecting to improve the overall performance of a computer by improving only one aspect of th
Oxana [17]

Answer:

a) For this case the new time to run the FP operation would be reduced 20% so that means 100-20% =80% from the original time

(1-0.2)*70 s =56s

The reduction on this case is 70-56 s=14s

And since the new total time would be given by 250-14=236 s

b) For this case the total time is reduced 20%  so that means that the new total time would be (1-0.2)=0.8 times the original total time (1-0.2) *250s =200 s

The original time for INT operations is calculated as:

250 = 70+85+40 +t_{INT}

t_{INT}=55s

For this part the only time that was changed is assumed the INT operations so then:

200 = 70+85+40 \Delta t_{INT}

And then: \Delta t_{INT}= 200-70-85-40=5 s

c) A reduction of the total time implies that the total time would be 205 s from the results above. And the time for FP is 70, for L/S is 85 and for INT operations is 55 s, so then if we add 70+85+55=210s, we see that 210>205 so then we cannot reduce the total time 20% just reducing the branch intructions.

Explanation:

From the info given we know that a computer running a program that requires 250 s, with 70 s spent executing FP instructions, 85 s executed L/S instructions and 40 s spent executing branch instructions.

Part 1

For this case the new time to run the FP operation would be reduced 20% so that means 100-20% =80% from the original time

(1-0.2)*70 s =56s

The reduction on this case is 70-56 s=14s

And since the new total time would be given by 250-14=236 s

Part 2

For this case the total time is reduced 20%  so that means that the new total time would be (1-0.2)=0.8 times the original total time (1-0.2) *250s =200 s

The original time for INT operations is calculated as:

250 = 70+85+40 +t_{INT}

t_{INT}=55s

For this part the only time that was changed is assumed the INT operations so then:

200 = 70+85+40 \Delta t_{INT}

And then: \Delta t_{INT}= 200-70-85-40=5 s

And we can quantify the decrease using the relative change:

\% Change = \frac{5s}{55 s} *100 = 9.09\% of reduction

Part 3

A reduction of the total time implies that the total time would be 205 s from the results above. And the time for FP is 70, for L/S is 85 and for INT operations is 55 s, so then if we add 70+85+55=210s, we see that 210>205 so then we cannot reduce the total time 20% just reducing the branch intructions.

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2 years ago
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Answer:

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3 years ago
what are the characteristics of an ideal fluid the general relation between shear stress and velocity gradient​
Dafna11 [192]

Answer:

ideal fluid follow Newtonian law

that is, shear stress is directly proportional to rate change of shear strain.

watch handwritten explanation

6 0
3 years ago
Problem 2. The length of a side of the square block is 4 in. Under the application of the load V, the top edge of the block disp
White raven [17]

Answer and Explanation:

The answer is attached below

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3 years ago
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