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>
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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
Answer:
2.145×10^-10 V or 0.2145nV
Explanation:
From hf=eV
h= Plank's constant = 6.6×10^-34JS
f= frequency of the electromagnetic wave = 5.2×10^4 Hz
e= electronic charge= 1.6×10^-19 C
V= voltage
V= hf/e
V= 6.6×10^-34JS × 5.2×10^4 Hz/ 1.6×10^-19 C
V= 2.145×10^-10 V or 0.2145nV
Therefore the voltage created is 2.145×10^-10 V or 0.2145nV
Answer:
1.) 274.5v
2.) 206.8v
Explanation:
1.) Given that In one part of the lab activities, students connected a 2.50 µF capacitor to a 746 V power source, whilst connected a second 6.80 µF capacitor to a 562 V source.
The potential difference and charge across EACH capacitor will be
V = Voe
Where Vo = initial voltage
e = natural logarithm = 2.718
For the first capacitor 2.50 µF,
V = Vo × 2.718
746 = Vo × 2.718
Vo = 746/2.718
Vo = 274.5v
To calculate the charge, use the below formula.
Q = CV
Q = 2.5 × 10^-6 × 274.5
Q = 6.86 × 10^-4 C
For the second capacitor 6.80 µF
V = Voe
562 = Vo × 2.718
Vo = 562/2.718
Vo = 206.77v
The charge on it will be
Q = CV
Q = 6.8 × 10^-6 × 206.77
Q = 1.41 × 10^-3 C
B.) Using the formula V = Voe again
165 = Vo × 2.718
Vo = 165 /2.718
Vo = 60.71v
Q = C × 60.71
Q = C
Answer:
x= 9.53 ounces
Explanation:
Given that
Mean ,μ= 9 ounces
Standard deviation ,σ=0.8 ounces
He wants to sell only those potatoes that are among the heaviest 25%.
P=25% = 0.25
When P= 0.25 then Z=0.674
Lest take x is the the minimum weight required to be brought to the farmer's market.
We know that
x = Z . σ + μ
x= 0.674 ₓ 0.8 + 9 ounces
x= 9.53 ounces
