The principal moments of inertia ( I max , I min ) for the angle’s cross-sectional area.

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Chapter 10.7, Problem 79P

To determine

The principal moments of inertia (Imax,Imin) for the angle’s cross-sectional area.

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USE MATLAB ONLY provide typed code solve for velocity triangle and dont provide copied answer Turbomachienery . GIven: vx = 185 m/s, flow angle = 60 degrees, R = 0.5, U = 150 m/s, b2 = -a3, a2 = -b3 Find: velocity triangle , a. magnitude of abs vel leaving rotor (m/s) b. flow absolute angles (a1, a2, a3) 3. flow rel angles (b2, b3) d. specific work done e. use code to draw vel. diagram Use this code for plot % plots Velocity Tri. in Ch4 function plotveltri(al1,al2,al3,b2,b3) S1L = [0 1]; V1x = [0 0]; V1s = [0 1*tand(al3)]; S2L = [2 3]; V2x = [0 0]; V2s = [0 1*tand(al2)]; W2s = [0 1*tand(b2)]; U2x = [3 3]; U2y = [1*tand(b2) 1*tand(al2)]; S3L = [4 5]; V3x = [0 0]; V3r = [0 1*tand(al3)]; W3r = [0 1*tand(b3)]; U3x = [5 5]; U3y = [1*tand(b3) 1*tand(al3)]; plot(S1L,V1x,'k',S1L,V1s,'r',... S2L,V2x,'k',S2L,V2s,'r',S2L,W2s,'b',U2x,U2y,'g',... S3L,V3x,'k',S3L,V3r,'r',S3L,W3r,'b',U3x,U3y,'g',...... 'LineWidth',2,'MarkerSize',10),... axis([-1 6 -4 4]), ...…

The answer should equal to 1157. Please sent me the solution. Thank you!

BONUS: If the volume of the 8cm x 6.5cm x 6cm Block of Aluminum was 312cm3 before machining, find how much material was removed when the fixture below was machined. +2 2.00 cm 6.00 cm 2.50 cm 6.50 cm 1.00 cm 2.50 cm 11.00 cm 8.00 cm 30 CP 9411 FL.4) (m² 1157 Area of triangle = 1/2*B*H Area of circle = лR² Circumference of a circle = 2πR 6.00 cm 6.50 cm 1.50 cm Radius 1.50 cm 1.00 cm