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
The car will take 300 m before it stops due to applying break.
<h3>What's the relation between initial velocity, final velocity, acceleration and distance?</h3>
- As per Newton's equation of motion, V² - U² = 2aS
- V= final velocity velocity of the object, U = initial velocity velocity of the object, a= acceleration, S = distance covered by the object
- Here, U = 60 ft/sec, V = 0 m/s, a= -6 ft/sec²
- So, 0² - 60² = 2×6× S
=> -3600 = -12S
=> S = 3600/12 = 300 m
Thus, we can conclude that the distance covered by the car is 300 m before it stopped.
Disclaimer: The question was given incomplete on the portal. Here is the complete question.
Question: A car is being driven at a rate of 60 ft/sec when the brakes are applied. The car decelerates at a constant rate of 6 ft/sec². How long will it take before the car stops?
Learn more about the Newton's equation of motion here:
brainly.com/question/8898885
#SPJ1
Answer:
jfjcgufnfhfufm TV fifnricnrhkddufnfif km fgkfkvntfmrugrhfifnh r
Answer:
What are we supposed to find, if it is kinetic energy then this is the solution.
K.E=1/2mv^2
K.E= kinetic energy
M=mass
V=velocity
K.E =0.5*55*0.6^2
K.E=9.9J
Explanation: