Answer:
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Answer:
29.16 J
Explanation:
From Hook's law,
W = 1/2(ke²)..................... Equation 1
Where W = work done, k = Spring constant, e = extension.
Given: W = 9 J, e = 0.5 m.
Substitute into equation 1
9 = 1/2(k×0.5²)
Solve for k
k = 18/0.5²
k = 72 N/m.
The work done required to stretch the spring by additional 0.4 m is
W = 1/2(72)(0.4+0.5)²
W = 36(0.9²)
W = 29.16 J.
Answer:
The angle between the red and blue light is 1.7°.
Explanation:
Given that,
Wavelength of red = 656 nm
Wavelength of blue = 486 nm
Angle = 37°
Suppose we need to find the angle between the red and blue light as it leaves the prism
We need to calculate the angle for red wavelength
Using Snell's law,
Put the value into the formula
We need to calculate the angle for blue wavelength
Using Snell's law,
Put the value into the formula
We need to calculate the angle between the red and blue light
Using formula of angle
Put the value into the formula
Hence, The angle between the red and blue light is 1.7°.
Answer:
a) The minimum thickness of the oil slick at the spot is 313 nm
b) the minimum thickness be now will be 125 nm
Explanation:
Given the data in the question;
a) The index of refraction of the oil is 1.20. What is the minimum thickness of the oil slick at that spot?
t = λ/2n
given that; wavelength λ = 750 nm and index of refraction of the oil n = 1.20
we substitute
t = 750 / 2(1.20)
t = 750 / 2.4
t = 312.5 ≈ 313 nm
Therefore, The minimum thickness of the oil slick at the spot is 313 nm
b)
Suppose the oil had an index of refraction of 1.50. What would the minimum thickness be now?
minimum thickness of the oil slick at the spot will be;
t = λ/4n
given that; wavelength λ = 750 nm and index of refraction of the oil n = 1.50
we substitute
t = 750 / 4(1.50)
t = 750 / 6
t = 125 nm
Therefore, the minimum thickness be now will be 125 nm