The distance mirror M2 must be moved so that one wavelength has produced one more new maxima than the other wavelength is;
<u><em>L = 57.88 mm</em></u>
<u><em /></u>
We are given;
Wavelength 1; λ₁ = 589 nm = 589 × 10⁻⁹ m
Wavelength 2; λ₂ = 589.6 nm = 589.6 × 10⁻⁹ m
We are told that L₁ = L₂. Thus, we will adopt L.
Formula for the number of bright fringe shift is;
m = 2L/λ
Thus;
For Wavelength 1;
m₁ = 2L/(589 × 10⁻⁹)
For wavelength 2;
m₂ = 2L/(589.6)
Now, we are told that one wavelength must have produced one more new maxima than the other wavelength. Thus;
m₁ - m₂ = 2
Plugging in the values of m₁ and m₂ gives;
(2L/589) - (2L/589.6) = 2
divide through by 2 to get;
L[(1/589) - (1/589.6)] = 1
L(1.728 × 10⁻⁶) = 1
L = 1/(1.728 × 10⁻⁶)
L = 578790.67 nm
L = 57.88 mm
Read more at; brainly.com/question/17161594
(a) The time for the capacitor to loose half its charge is 2.2 ms.
(b) The time for the capacitor to loose half its energy is 1.59 ms.
<h3>
Time taken to loose half of its charge</h3>
q(t) = q₀e-^(t/RC)
q(t)/q₀ = e-^(t/RC)
0.5q₀/q₀ = e-^(t/RC)
0.5 = e-^(t/RC)
1/2 = e-^(t/RC)
t/RC = ln(2)
t = RC x ln(2)
t = (12 x 10⁻⁶ x 265) x ln(2)
t = 2.2 x 10⁻³ s
t = 2.2 ms
<h3>
Time taken to loose half of its stored energy</h3>
U(t) = Ue-^(t/RC)
U = ¹/₂Q²/C
(Ue-^(t/RC))²/2C = Q₀²/2Ce
e^(2t/RC) = e
2t/RC = 1
t = RC/2
t = (265 x 12 x 10⁻⁶)/2
t = 1.59 x 10⁻³ s
t = 1.59 ms
Thus, the time for the capacitor to loose half its charge is 2.2 ms and the time for the capacitor to loose half its energy is 1.59 ms.
Learn more about energy stored in capacitor here: brainly.com/question/14811408
#SPJ1
True
The half-life isn’t applicable to a first order reaction because it does not rely on the concentration of reactant present. However the 2nd order reaction is dependent on the concentration of the reactant present.
The relationship between the half life and the reactant is an inverse one.
The half life is usually reduced or shortened with an increase in the concentration and vice versa.
CaS is the empirical formula of the compound between calcium and sulphur that has the percent composition 55.6.
When percentages are given, we take a total mass of 100 grams.
Therefore the mass of each element is equal to the percentage given.
Mass of Ca = 55.6 g (given) of
S Mass = 44.4 g (100 - 55.6 = 44.4)
Step 1: Convert the given mass to moles.
moles Ca = given mass Ca / molar mass Ca
moles = 55.6 / 40 = 1.39 moles
mol S = specific mass S / molar mass S
mol = 44.4 / 32 = 1.39 mol
Step 2: Divide the molar ratio of each molar value by the smallest number of moles calculated.
For Ca = 1.39 / 1.39 = 1
For S = 1.39 / 1.39 = 1
The ratio of Ca : S = 1:1
Hence the empirical formula of the given compound will be CaS.
Learn more about empirical formula here : brainly.com/question/1496676
#SPJ4
