Actually moving and not. It is the sum of potential and kinetic energy.
Answer:
4.981 MeV
Explanation:
The quantity of energy Q can be calculated using the formula
Q = (mass before - mass after) × c²
Atomic Mass of thorium = 232.038054 u, atomic of Radium = 228.0301069 u and mass of Helium = 4.00260. The difference of atomic number and atomic mass between the thorium and radium ( 232 - 228) and ( 90 - 88) show α particle was emitted.
1 u = 931.494 Mev/c²
Q = (mass before - mass after) × c²
Q = ( mass of thorium - ( mass of Radium + mass of Helium ) )× c²
Q = 232.038054 u - ( 228.0301069 + 4.00260) × c²
Q = 0.0053471 u × c²
replace 1 u = 931.494 MeV/ c²
Q = 0.0053471 × c² × (931.494 MeV / c²)
cancel c² from the equation
Q = 0.0053471 × 931.494 MeV = 4.981 MeV
We will first record its mass and then its volume by measuring its dimensions
then divide mass by volume and will get density of regular solid
Answer:
The force constant is 
The energy stored in the spring is 
Explanation:
From the question we are told that
The mass of the object is 
The period is 
The period of the spring oscillation is mathematically represented as
where k is the force constant
So making k the subject
substituting values
The energy stored in the spring is mathematically represented as
Where x is the spring displacement which is given as
substituting values
Answer:
1.635×10^-3m
Explanation:
Young modulus is the ratio of the tensile stress of a material to its tensile strain.
Young modulus = Tensile stress/tensile strain
Tensile stress = Force/Area
Given force = 130N
Area = Πr² = Π×(1.55×10^-3)²
Area = 4.87×10^-6m²
Tensile stress = 130/4.87×10^-6 = 8.39×10^7N/m²
Tensile strain = extension/original length
Tensile strain = e/3.9
Substituting in the young modulus formula given young modulus to be 2×10¹¹N/m²
2×10¹¹N/m² = 8.39×10^7/{e/3.9)}
2×10¹¹ = (8.39×10^7×3.9)/e
2×10¹¹e = 3.27×10^8
e = 3.27×10^8/2×10¹¹
e = 1.635×10^-3m
The stretch of the steel wire will be
1.635×10^-3m
