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
Answer:
a= 0.22 m/s²
Explanation:
Given that
M = 3.5 kg
θ = 30°
m = 1 kg
μ= 0.3
The force due to gravity
F₁= M g sinθ
F₁=3.5 x 10 x sin 30
F₁= 17.5 N
F₂ = m g
F₂ = 1 x 10 = 10 N
The maximum value of the friction force on the incline plane
Fr = μ M g cosθ
Fr = 0.3 x 2.5 x 10 cos30°
Fr= 6.49 N
Lets take acceleration of the system is a m/s²
F₁ - F₂ - Fr = (M+m) a
17.5 - 10 - 6.49 = (3.5+1)a
a= 0.22 m/s²
There are 20 joules of useful energy.
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