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3 years ago

Question 54 (1 point)

Engineering

2 answers:

Rudiy273 years ago

6 0

The answer should be Oc 15

Sergeeva-Olga [200]3 years ago

6 0

Oa 8 should be the answer of the moisture of lumber frame for the house

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1. A gas pressure difference is applied to the legs of a U-tube manometer filled with a liquid with S = 1.5. The manometer readi

julia-pushkina [17]

Answer:

1) The pressure difference is 4.207 kilopascals.

2) 2.5 pounds per square inch equals 5.093 inches of mercury and 5.768 feet of water.

Explanation:

1) We can calculate the gas pressure difference from the U-tube manometer by using the following hydrostatic formula:

$\Delta P = \frac{S\cdot \rho_{w}\cdot g \cdot \Delta h}{1000}$ (Eq. 1)

Where:

$S$ - Relative density, dimensionless.

$\rho_{w}$ - Density of water, measured in kilograms per cubic meter.

$g$ - Gravitational acceleration, measured in meters per square second.

$\Delta h$ - Height difference in the U-tube manometer, measured in meters.

$\Delta P$ - Gas pressure difference, measured in kilopascals.

If we know that $S = 1.5$ , $\rho_{w} = 1000\,\frac{kg}{m^{3}}$ , $g = 9.807\,\frac{m}{s^{2}}$ and $\Delta h = 0.286\,m$ , then the pressure difference is:

$\Delta P = \frac{1.5\cdot \left(1000\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.286\,m)}{1000}$

$\Delta P = 4.207\,kPa$

The pressure difference is 4.207 kilopascals.

2) From Physics we remember that a pound per square unit equals 2.036 inches of mercury and 2.307 feet of water and we must multiply the given pressure by corresponding conversion unit: ( $p = 2.5\,psi$ )

$p = 2.5\,psi\times 2.037\,\frac{in\,Hg}{psi}$

$p = 5.093\,in\,Hg$

$p = 2.5\,psi\times 2.307\,\frac{ft\,H_{2}O}{psi}$

$p = 5.768\,ft\,H_{2}O$

2.5 pounds per square inch equals 5.093 inches of mercury and 5.768 feet of water.

4 0

4 years ago

Ok there........................................................................

Juliette [100K]

Answer:

ok THERE

Explanation:

4 0

3 years ago

Suppose there are 93 packets entering a queue at the same time. Each packet is of size 4 MiB. The link transmission rate is 1.4

Ghella [55]

Answer:

0.19s

Explanation:

Queueing delay is the time a job waits in a queue before it can be executed. it is the difference in time betwen when the packet data reaches it destination and the time when it was executed.

Queueing delay =(N-1) L /2R

where N = no of packet =93

L = size of packet = 4MB

R = bandwidth = 1.4Gbps = 1×10⁹ bps

4 MB = 4194304 Bytes

(93 - 1)4194304 / 2× 10⁹

queueing delay =192937984 ×10⁻⁹

=0.19s

5 0

3 years ago

Type the correct answer in the box. Spell all words correctly. Elaine is an engineer. She has to create a prototype of a machine

Trava [24]

Answer:

the mean time I was wondering if you have any questions or need any further information please contact me at the mean time I was

8 0

3 years ago

Implement the function lastChars() that takes a list of strings as a parameter and prints to the screen the last character of ea

Liono4ka [1.6K]

Answer:

The following program is in C++.

#include <bits/stdc++.h>

using namespace std;

void lastChars(string s)

{

int l=s.length();

if(l!=0)

{

cout<<"The last character of the string is: "<<s[l-1];

}

int main() {

string s;//declaring a string..

getline(cin,s);//taking input of the string..

lastChars(s);//calling the function..

return 0;

}

Input:-

Alex is going home

Output:-

The last character of the string is: e

Explanation:

In the function lastChars() there is one argument that is a string.I have declared a integer variable l that stores the length of the string.If the length of the string is not 0.Then printing the last character of the string.In the main function I have called the function lastChars() with the string s that is prompted from the user.

8 0

3 years ago