All categories

1 year ago

How high of a column of sae 30 oil would be required to give the same pressure as 700 mm hg?

Engineering

1 answer:

Rasek [7]1 year ago

7 0

Hoiu-10,4000 mm.

<h3>Is positive pressure good for PC?</h3>

A balanced configuration is the most efficient way to cool your pc although it should tend towards a slight positive pressure if you can help it.
Tip: As much as it might seem important, the concept of heat rising doesn't have too much of an effect.

To learn more about it, refer

to https://brainly.in/question/413163

#SPJ4

You might be interested in

Using the following data, determine the percentage retained, cumulative percentage retained, and percent passing for each sieve.

vekshin1

Solution :

<u>Sieve Size</u> (in) <u>Weight retain</u><u>(g)</u>

3 1.62

2 2.17

$1\frac{1}{2}$ 3.62

$\frac{3}{4}$ 2.27

$\frac{3}{8}$ 1.38

PAN 0.21

Given :

Sieve weight % wt. retain % cumulative % finer

size retained wt. retain

No. 4 59.5 10.225% 10.225% 89.775%

No. 8 86.5 14.865% 25.090% 74.91%

No. 16 138 23.7154% 48.8054% 51.2%

No. 30 127.8 21.91% 70.7154% 29.2850%

No. 50 97 16.6695% 87.3849% 12.62%

No. 100 66.8 11.4796% 98.92% 1.08%

Pan <u> 6.3 </u> 1.08% 100% 0%

581.9 gram

Effective size = percentage finer 10% ( $$$D_{20}$ )

0.149 mm, N 100, % finer 1.08

0.297, N 50 , % finer 12.62%

x , 10%

$y-1.08 = \frac{12.62 - 1.08}{0.297 - 0.149}(x-0.149)$

$(10-1.08) \times \frac{0.297 - 0.149}{12.62 - 1.08}+ 0.149=x$

x = 0.2634 mm

Effective size, $D_{10} = 0.2643 \ mm$

Now, N 16 (1.19 mm) , 51.2%

N 8 (2.38 mm) , 74.91%

x, 60%

$60-51.2 = \frac{74.91-51.2}{2.38-1.19}(x-1.19)$

x = 1.6317 mm

$\therefore D_{60} = 1.6317 \ mm$

Uniformity co-efficient = $\frac{D_{60}}{D_{10}}$

$Cu= \frac{1.6317}{0.2643}$

Cu = 6.17

Now, fineness modulus = $\frac{\Sigma \text{\ cumulative retain on all sieve }}{100}$

$=\frac{\Sigma (10.225+25.09+48.8054+70.7165+87.39+98.92+100)}{100}$

= 4.41

which lies between No. 4 and No. 5 sieve [4.76 to 4.00]

So, fineness modulus = 4.38 mm

7 0

3 years ago

If you are setting up a race car. What is the cross weight? Does it matter?

lara31 [8.8K]

Answer:

cross-weight is used to tighten it up.

Explanation:

and yes this is important because Cross-weight percentage compares the diagonal weight totals to the car's total weight.

hope this help

(mark this answer as an brainliest answer)

7 0

4 years ago

Explain what a margin of safety is in driving as well as how it can help minimize risk.

Yakvenalex [24]

Answer:

A safety margin is the space left between your vehicle and the next to provide room, time and visibility at every instant

Explanation:

A safety margin is defined as an allowance given between your vehicle and the next vehicle in front to provide enough room, visibility and time to move in a safe manner to prevent the occurrence of an accident at anytime the frontal vehicle suddenly stops or slows down

Safety margins help minimize risks in the following way

1) A common knowledge of safety margins, improves predictability among road users, thereby minimizing the risk traffic accidents caused due to late communication

2) The use of safety margins helps minimize the risk due to a change in driving conditions such as when the road becomes more slippery from being covered with fluid that is being wetted

3) Safety margin can help prevent the occurrence of an accident between vehicles due to failure of a car system, such as a punctured tire or failed breaking system

4) Safety margin helps to protect road users from the introduction of obstacles on the main roads such as ongoing road construction, broken down vehicles, road blockage by vehicles involved in an accident etc

5) Safety margin help protect road users from being involved in an accident due to the loss of driving focus of the driver of the frontal vehicle

6 0

3 years ago

A 14 inch diameter pipe is decreased in diameter by 2 inches through a contraction. The pressure entering the contraction is 28

Delicious77 [7]

Answer:

5984.67N

Explanation:

A 14 inch diameter pipe is decreased in diameter by 2 inches through a contraction. The pressure entering the contraction is 28 psi and a pressure drop of 2 psi occurs through the contraction if the upstream velocity is 4.0 ft/sec. What is the magnitude of the resultant force (lbs) needed to hold the pipe in place?

from continuity equation

v1A1=v2A2

equation of continuity

v1=4ft /s=1.21m/s

d1=14 inch=.35m

d2=14-2=0.304m

A1=pi*d^2/4

0.096m^2

a2=0.0706m^2

from continuity once again

1.21*0.096=v2(0.07)

v2=1.65