Answer:
and my favorite song is popular loner
Explanation:
my favorite rapper is rod wave
Answer:
C. 14.55
Explanation:
12 x 10 = 120
120 divded by 10 is 12
so now we do the left side
7 x 3 = 21 divded by 10 is 2
so now we have 14
and the remaning area is 0.55
so 14.55
Solution :
Given :
External diameter of the hemispherical shell, D = 500 mm
Thickness, t = 20 mm
Internal diameter, d = D - 2t
= 500 - 2(20)
= 460 mm
So, internal radius, r = 230 mm
= 0.23 m
Density of molten metal, ρ = 
= 
The height of pouring cavity above parting surface is h = 300 mm
= 0.3 m
So, the metallostatic thrust on the upper mold at the end of casting is :

Area, A 




= 7043.42 N
Answer:
The power developed in HP is 2702.7hp
Explanation:
Given details.
P1 = 150 lbf/in^2,
T1 = 1400°R
P2 = 14.8 lbf/in^2,
T2 = 700°R
Mass flow rate m1 = m2 = m = 11 lb/s Q = -65000 Btu/h
Using air table to obtain the values for h1 and h2 at T1 and T2
h1 at T1 = 1400°R = 342.9 Btu/h
h2 at T2 = 700°R = 167.6 Btu/h
Using;
Q - W + m(h1) - m(h2) = 0
W = Q - m (h2 -h1)
W = (-65000 Btu/h ) - 11 lb/s (167.6 - 342.9) Btu/h
W = (-65000 Btu/h ) - (-1928.3) Btu/s
W = (-65000 Btu/h ) * {1hr/(60*60)s} - (-1928.3) Btu/s
W = -18.06Btu/s + 1928.3 Btu/s
W = 1910.24Btu/s
Note; Btu/s = 1.4148532hp
W = 2702.7hp