Answer:
The shear strain is 0.05797 rad.
Explanation:
Shear strain is the ratio of change in dimension along the shearing load direction to the height of the plate under application of shear load. Width of the plate remains same. Length of the plate slides under shear load.
Step1
Given:
Height of the pad is 1.38 in.
Deformation at the top of the pad is 0.08 in.
Calculation:
Step2
Shear strain is calculated as follows:



For small angle of
,
can take as
.

Thus, the shear strain is 0.05797 rad.
They ran different shapes and materials through a wind tunnel to see which shape and material would decrease energy output so that it takes in equal COthan it puts out.
Answer:
Enthalpy at outlet=284.44 KJ
Explanation:


We need to Find enthalpy of outlet.
Lets take the outlet mass m and outlet enthalpy h.
So from mass conservation

m=1+1.5+2 Kg/s
m=4.5 Kg/s
Now from energy conservation

By putting the values

So h=284.44 KJ
Answer:
elongation of the brass rod is 0.01956 mm
Explanation:
given data
length = 5 cm = 50 mm
diameter = 4.50 mm
Young's modulus = 98.0 GPa
load = 610 N
to find out
what will be the elongation of the brass rod in mm
solution
we know here change in length formula that is express as
δ =
................1
here δ is change in length and P is applied load and A id cross section area and E is Young's modulus and L is length
so all value in equation 1
δ =
δ =
δ = 0.01956 mm
so elongation of the brass rod is 0.01956 mm