Answer:
Part a)

Part b)

Explanation:
Part A)
As we know that time period of the motion is given as

so we have



now at the point of maximum amplitude the force equation when Normal force is about to zero is given as

so we have



Part b)
Now if the amplitude of the SHM is 6.23 cm
and now at this amplitude if object will lose the contact then in that case again we have



so now we have


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.
Answer:
hi apner how is ur going for the weekend on your way home
Explanation:
hi there is that something you could help me
Answer:
100
Explanation:
take note that v=d/t (velocity is distance over(divided by) time, so in this case it would be 200 (distance) divided by 2 (time) = 100
Answer:
The box of rocks will have depression which can be seen without touching the box.
Explanation:
The density of rocks is very large as compared with napkins. So, the weight of the rocks will be much more greater than that of napkins.
As both boxes have same volume the heavier box will show depression on the lower surface as compared to the lighter box. So, the box of rocks will have depression which can be seen without touching the box.