For this, we must know the density of lead. This is 11.36 grams per cubic centimeter. The mass present is 408 grams. Therefore, the volume occupied will be:
408 / 11.36 = 35.9 cubic centimeters. This is the volume of water that will be displaced.
Now, we know that one cubic centimeter is equivalent to one mL. Therefore,
35.9 ml of water will be displaced.
I am sorry if I am wrong but, the net force would be zero. 0
<span>Assume: neglect of the collar dimensions.
Ď_h=(P*r)/t=(5*125)/8=78.125 MPa ,Ď_a=Ď_h/2=39 MPa
τ=(S*Q)/(I*b)=(40*〖10〗^3*π(〖0.125〗^2-〖0.117〗^2 )*121*〖10〗^(-3))/(π/2 (〖0.125〗^4-〖0.117〗^4 )*8*〖10〗^(-3) )=41.277 MPa
@ Point K:
Ď_z=(+M*c)/I=(40*0.6*121*〖10〗^(-3))/(8.914*〖10〗^(-5) )=32.6 MPa
Using Mohr Circle:
Ď_max=(Ď_h+Ď_a)/2+âš(Ď„^2+((Ď_h-Ď_a)/2)^2 )
Ď_max=104.2 MPa, Ď„_max=45.62 MPa</span>
A..........................................
W, because as time is moving up at a consistent rate the speed is as well, creating the straight line.
