E = ½KA^2 is the mechanical energy of any oscillator. It is the sum of elastic potential energy and
kinetic energy. When amplitude A
decreases by 3%, then
(E2-E1)/E1 = {½K(A2^2/A1^2) }/ ½K(A1^2)
= {(A2^2 – A1^2) / (A1^2)}
= 97^2 – 100^2/100^2
= 5.91% of the mechanical energy is lost each cycle.
Elastic potential = 1/2 x constant x square of compression lenght
So it's 360 N/m
Answer:
0.056 psi more pressure is exerted by filled coat rack than an empty coat rack.
Explanation:
First we find the pressure exerted by the rack without coat. So, for that purpose, we use formula:
P₁ = F/A
where,
P₁ = Pressure exerted by empty rack = ?
F = Force exerted by empty rack = Weight of Empty Rack = 40 lb
A = Base Area = 452.4 in²
Therefore,
P₁ = 40 lb/452.4 in²
P₁ = 0.088 psi
Now, we calculate the pressure exerted by the rack along with the coat.
P₂ = F/A
where,
P₂ = Pressure exerted by rack filled with coats= ?
F = Force exerted by filled rack = Weight of Filled Rack = 65 lb
A = Base Area = 452.4 in²
Therefore,
P₂ = 65 lb/452.4 in²
P₂ = 0.144 psi
Now, the difference between both pressures is:
ΔP = P₂ - P₁
ΔP = 0.144 psi - 0.088 psi
<u>ΔP = 0.056 psi</u>
Answer:

Explanation:
The fusion reaction in this problem is

The total energy released in the fusion reaction is given by

where
is the speed of light
is the mass defect, which is the mass difference between the mass of the reactants and the mass of the products
For this fusion reaction we have:
is the mass of one nucleus of hydrogen
is the mass of one nucleus of helium
So the mass defect is:

The conversion factor between atomic mass units and kilograms is

So the mass defect is

And so, the energy released is:
