Answer:
Explanation:
From the information given:
The total load is distributed across both the rod and tube:
Since this is a composite column; the elongation of both aluminum rod & steel tube is equal.
Replace 
into equation (1)
Finally, to determine the normal stress in aluminum rod:
Thus, the normal stress = 23.523 MPa in compression.
Answer:
true
Explanation:
A well designed product will increase in sells and in stock.
Answer: the mass flow rate of concentrated brine out of the process is 46,666.669 kg/hr
Explanation:
F, W and B are the fresh feed, brine and total water obtained
w = 2 x 10^4 L/h
we know that
F = W + B
we substitute
F = 2 x 10^4 + B
F = 20000 + B .................EQUA 1
solute
0.035F = 0.05B
B = 0.035F/0.05
B = 0.7F
now we substitute value of B in equation 1
F = 20000 + 0.7F
0.3F = 20000
F = 20000/0.3
F = 66666.67 kg/hr
B = 0.7F
B = 0.7 * F
B = 0.7 * 66666.67
B = 46,666.669 kg/hr
the mass flow rate of concentrated brine out of the process is 46,666.669 kg/hr
Answer:
$$\begin{align*}
P(Y−X=m|Y>X)=∑kP(Y−X=m,X=k|Y>X)=∑kP(Y−X=m|X=k,Y>X)P(X=k|Y>X)=∑kP(Y−k=m|Y>k)P(X=k|Y>X).
Explanation:
P(Y−X=m|Y>X)=∑kP(Y−X=m,X=k|Y>X)=∑kP(Y−X=m|X=k,Y>X)P(X=k|Y>X)=∑kP(Y−k=m|Y>k)P(X=k|Y>X).
