In a double-slit interference experiment, the distance y of the maximum of order m from the center of the observed interference pattern on the screen is

where D=5.00 m is the distance of the screen from the slits, and

is the distance between the two slits.
The fringes on the screen are 6.5 cm=0.065 m apart from each other, this means that the first maximum (m=1) is located at y=0.065 m from the center of the pattern.
Therefore, from the previous formula we can find the wavelength of the light:

And from the relationship between frequency and wavelength,

, we can find the frequency of the light:
From the Hooke's law , the extension force of an elastic material is directly proportional to the extension.
That is, F = k e, where F is the force , k is the constant and e is the extension
F = 10 × 10 = 100 N
e = 1mm or 0.001 m
Hence, k = F/e
= 100 N/ 0.001
= 100000 N/m or 100 N/mm
Answer:
volume is the correct answer
Explanation:
D=m/V therefore the answer is 120/200 or 0.6
For part a)
Since the conical surface is not exposed to the radiation coming from the walls only from the circular plate and assuming steady state, the temperature of the conical surface is also equal to the temperature of the circular plate. T2 = 600 K
For part b)
To maintain the temperature of the circular plate, the power required would be calculated using:
Q = Aσ(T₁⁴ - Tw⁴)
Q = π(500x10^-3)²/4 (5.67x10^-8)(600⁴ - 300⁴)
Q = 5410.65 W