Answer:
Here Strain due to testing is greater than the strain due to yielding that is why computation of load is not possible.
Explanation:
Given that
Yield strength ,Sy= 240 MPa
Tensile strength = 310 MPa
Elastic modulus ,E= 110 GPa
L=380 mm
ΔL = 1.9 mm
Lets find strain:
Case 1 :
Strain due to elongation (testing)
ε = ΔL/L
ε = 1.9/380
ε = 0.005
Case 2 :
Strain due to yielding
ε '=0.0021
Here Strain due to testing is greater than the strain due to yielding that is why computation of load is not possible.
For computation of load strain due to testing should be less than the strain due to yielding.
A perfect black body can’t be realized
Answer:
A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.
Explanation:
The bulk modulus is represented by the following differential equation:
Where:
- Bulk module, measured in pascals.
- Sample volume, measured in cubic meters.
- Local pressure, measured in pascals.
Now, let suppose that bulk remains constant, so that differential equation can be reduced into a first-order linear non-homogeneous differential equation with separable variables:
This resultant expression is solved by definite integration and algebraic handling:
The final volume is predicted by:
If 
, 
and 
, then:
Change in volume due to increasure on pressure is:
A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.
Answer:
it loaded and it is C. buddy sorry about that :)
Pretty sure it’s A. Hope this helps.
