Answer:
his type of shock is called inelastic
Explanation:
This exercise is for vehicle crashes, which corresponds to exercise is momentum conservation.
We must begin by defining a system formed by the two cars so that the forces during the crash have been intense and the moment is preserved.
Looking for the moments
initial. Before the crash
p₀ = m₁ v₁₀
final. After the crash
p_f = (m₁ + m₂) v
the conservation of the moment is written
p₀ = p_f
m₁ v₁₀ = (m₁ + m₂) v
This type of shock is called inelastic and has the characteristics that the kinetic energy is not conserved.
Answer:
strain = 0.0011575
Explanation:
Given:
- The diameter of specimen d = 0.02 m
- Applied uni-axial force P = 40,000 N
- Modulus of Elasticity E = 110 GPa
Find:
If the deformation is totally elastic, what is the strain experienced by the specimen?
Solution:
- First we need to calculate the stress in the specimen using the relation:
stress = Force / Area
stress = 40,000*4 / pi*d^2
stress = 160,000 / pi*0.02^2
stress = 127.323 MPa
- Now compute Elastic strain from the definition of Elastic modulus of a material:
E = stress / strain
strain = stress / E
- Plug in the values:
strain = 127.323 / 110*10^3
strain = 0.0011575
Let's know the formula
So
Hence if length is doubled time period will be √2 times
#2
Decrease by 5% will decrease the time period 5/2=2.5%
So
It's
- 100-2.5=97.5% of old period
Maintenance spanner are needed in great numbers to service all sorts of technical equipment
Answer:
After 5 half life the amount remains as 3.125 g
Explanation:
After each half life the quantity becomes half
so we know that initial amount is 100 g
so after 1st half life amount is half so it left as
m = 50 g
now after 2nd half life amount again becomes half
m = 25 g
after 3rd half life it is again half so remaining amount is
m = 12.5 g
after 4th half life the amount becomes half again
m = 6.25 g
Now after 5th half life the amount remains as
m = 3.125 g