J.K

The C.M of the system is closer to the heavier mass 5kg. Assume the C.M is a distance d from Y. mass 2kg is (5+d)m from c.m of system.5kg is dm from c.m and 4kg is (7-d)m from c.m taking moment about c.m:2(5+d)+5d=4(7-d) 10+2d+5d=28-4d 11d=18 d=1.64m c.m...

Parallel line has same gradient,m.from equation of line,m=4 y-y1=m(x-x1) y-2=4(x-8) y-2=4x-32 y=4x-30 b)m2=-1/m1=-1/4 y-2=-1/4(x-8) 4y-8=-x+8 4y=-x+16

According to work energy theorem workdone=change in kinetic energy of the body.W=1/2M(V2^2-V1^2)=1/2(38)(1.5^2-0.4^2)=39.71Joules

The two blocks decelerate due to friction Ffr=(m1+m2)a a=Ffr/(m1+m2) Ffr=u(m1+m2)g a=u(m1+m2)g/(m1+m2) cancelling the common parameter,a=ug a=0.39*10=3.9m/s Vf=sqrt(2*3.9*7.47)=7.63m/s b)150v1=(150+265)*7.63 v1=21.1m/s

W=18*10N=180N B-W=Ma B-180N=(18)(1.8) B=32.4+180=212.4N

W1/g1=W2/g2 1370/9.8=W2/25.9 W2=3621N

R=1/2h Assume the water in cone is at height,h and radius is r.V=1/3Pi(1/2h)^2(h) V=1/12(Pi)h^3. by differentiating V:dV/dt=1/4(pi)h^2dh/dt 10=1/4*3.142*8^2dh/dt dh/dt=10/50.27=0.199ft/min

At equilibrium,p1=p2 q2+12q+700=1800-10q-q2 2q2+22q-1100=0 q2+11q-550=0 solving q=18.59 p=18.59^2+12(18.59)+700=N1268.67

9v-(4*2)=0 9v=8 v=0.89m/s