To solve this problem, we will start by defining each of the variables given and proceed to find the modulus of elasticity of the object. We will calculate the deformation per unit of elastic volume and finally we will calculate the net energy of the system. Let's start defining the variables
Yield Strength of the metal specimen
Yield Strain of the Specimen
Diameter of the test-specimen
Gage length of the Specimen
Modulus of elasticity
Strain energy per unit volume at the elastic limit is
Considering that the net strain energy of the sample is
Therefore the net strain energy of the sample is 
It is (A) for the first one and (B) for the second one.
Answer:
the magnitude of the force that one particle exerts on the other is 79.08 N
Explanation:
given information:
q₁ = 3.77 μC = -3.77 x 10⁻⁶ C
q₂ = 4.39 μC = 4.39 x 10⁻⁶ C
r = 4.34 cm = 4.34 x 10⁻² m
What is the magnitude of the force that one particle exerts on the other?
lFl = kq₁q₂/r²
= (9 x 10⁹) (3.77 x 10⁻⁶) (4.39 x 10⁻⁶)/(4.34 x 10⁻²)²
= 79.08 N
Answer:
Work done= Energy transferred
Explanation:
Work is the transfer of energy. In physics we say that work is done on an object when you transfer energy to that object. If you put energy into an object, then you do work on that object (mass).
Answer:
For carbon 12
ΔV1 = 2265.31 V
ΔV2 = 362.5 V
For carbon 14
ΔV1 = 1941.7 V
ΔV2 = 310.67 V
Explanation:
The complete explanations are given in the attachment below. The formulae for the accelerating potential ΔV1 and ΔV2 are derived and the necessary parameters are substituted into the derived equations.
