Answer:
aluminum bar carrying a higher load than steel bar
Explanation:
Given data;
steel abr
diameter = 5 mm
stress = 500 MPa
aluminium bar
diameter = 10 mm
stress = 150 MPa
we know
stress = laod/area
for steel bar
solving for P
P = 9817.47 N
for Aluminium bar
solving for P
P = 11790 N
aluminum bar carrying a higher load than steel bar
Using the Bilinear Transformation, determine the order N and the cut off frequency c of the analog prototype filter for the following discrete time design: a) passband
Answer:
shearing deformation has to do with change in shape not length, so it should be neglected. As for the data obtained in this experiment, it strongly agree with the statement as all values in the experiment were quite close and were under the target value. Greater than ten percent error.
Answer:
0.124
Explanation:
We calculate the hydraulic gradient by the formulas below.
I = (change in h)/(change in l)-----eqn 1
I = (hk-hl)/change in L ----- equation 2
At k the headloss = hk,
At L the headloss = hL
The distance of water travel is change in I
Total head at k
hk = 543+23
= 566 ft
Total head at L
hL = 461+74
= 535 ft
Change in L = 250
When we substitute these values in equation 2
566-535/250
= 0.124
The hydraulic gradient is 0.124