Answer:
no
Explanation:
it's not a dead load because when load is put on the pillars it's not fully straining it's been slowly getting to be heavier in that period of time before it falls
Answer:
Heat required =7126.58 Btu.
Explanation:
Given that
Mass m=20 lb
We know that
1 lb =0.45 kg
So 20 lb=9 kg
m=9 kg
Ice at -15° F and we have to covert it at 200° F.
First ice will take sensible heat at up to 32 F then it will take latent heat at constant temperature and temperature will remain 32 F.After that it will convert in water and water will take sensible heat and reach at 200 F.
We know that
Specific heat for ice 
Latent heat for ice H=336 KJ/kg
Specific heat for ice 
We know that sensible heat given as
Heat for -15F to 32 F:
Q=858.69 KJ
Heat for 32 Fto 200 F:
Q=6330.74 KJ
Total heat=858.69 + 336 +6330.74 KJ
Total heat=7525.43 KJ
We know that 1 KJ=0.947 Btu
So 7525.43 KJ=7126.58 Btu
So heat required to covert ice into water is 7126.58 Btu.
Answer:
<u>The automobile rental prices shall show all taxes (including a 6% state tax).</u>
Explanation:
Im pretty sure
Answer:
M2 = 0.06404
P2 = 2.273
T2 = 5806.45°R
Explanation:
Given that p1 = 10atm, T1 = 1000R, M1 = 0.2.
Therefore from Steam Table, Po1 = (1.028)*(10) = 10.28 atm,
To1 = (1.008)*(1000) = 1008 ºR
R = 1716 ft-lb/slug-ºR cp= 6006 ft-lb/slug-ºR fuel-air ratio (by mass)
F/A =???? = FA slugf/slugaq = 4.5 x 108ft-lb/slugfx FA slugf/sluga = (4.5 x 108)FA ft-lb/sluga
For the air q = cp(To2– To1)
(Exit flow – inlet flow) – choked flow is assumed For M1= 0.2
Table A.3 of steam table gives P/P* = 2.273,
T/T* = 0.2066,
To/To* = 0.1736 To* = To2= To/0.1736 = 1008/0.1736 = 5806.45 ºR Gives q = cp(To* - To) = (6006 ft-lb/sluga-ºR)*(5806.45 – 1008)ºR = 28819500 ft-lb/slugaSetting equal to equation 1 above gives 28819500 ft-lb/sluga= FA*(4.5 x 108) ft-lb/slugaFA =
F/A = 0.06404 slugf/slugaor less to prevent choked flow at the exit
