Answer:
function summedValue = SummationWithLoop(userNum)
% Summation of all values from 1 to userNum
summedValue = 0;
i = 0;
% use a while loop that assigns summedValue with the
% sum of all values from 1 to userNum
while(i <= userNum)
summedValue = summedValue + i;
i = i + 1;
end
end
Answer:
b)Poly crystalline and amorphous materials with small diameter
Explanation:
Fibers have high length to diameter ratio and also have high strength.Generally length of fibers is very high and diameter is very low as compare to length.
Mostly fibers is used to transfer data from one place to another place with help of fiber optical cables.Fiber optic cables is used in telecommunication.In these cables data covert in to the electric single and reach at define location and after data is decode and covert from electric single in to original data.
Fibers poly crystalline and amorphous materials with small diameter.
Answer:
One of the reasons why flashover fires are more prevalent today than it was in the past is that homes and furniture today are made from materials that are far more combustible than those of previous years.
Explanation:
A flashover fire is the rapid ignition and combustion of all flammable materials in an enclosed vicinity in a very short period of time.
Thirty years ago, the average escape time from a house that was on fire is about sixteen and fifty seconds...that would be approximately seventeen minutes. Presently that figure is down to four minutes.
One of the reasons identified is that the internal and external appurtenances especially furniture in use today are more combustible than those of previous years. That is, as they burn, they produce more heat and disintegrate faster.
The reason identified for this is, old houses were made of more natural materials such as real wood etc whilst the furniture and curtains in modern houses are mostly from synthetic materials.
Cheers
Answer:64.10 Btu/lbm
Explanation:
Work done in an isothermally compressed steady flow device is expressed as
Work done = P₁V₁ In { P₁/ P₂}
Work done=RT In { P₁/ P₂}
where P₁=13 psia
P₂= 80 psia
Temperature =°F Temperature is convert to °R
T(°R) = T(°F) + 459.67
T(°R) = 55°F+ 459.67
=514.67T(°R)
According to the properties of molar gas, gas constant and critical properties table, R which s the gas constant of air is given as 0.06855 Btu/lbm
Work = RT In { P₁/ P₂}
0.06855 x 514.67 In { 13/ 80}
=0.06855 x 514.67 In {0.1625}
= 0.06855 x 514.67 x -1.817
=- 64.10Btu/lbm
The required work therefore for this isothermal compression is 64.10 Btu/lbm
