This is a Fraunhofer single slit experiment, where the light passing through the slit produces an interference pattern on the screen, and where the dark bands (minima of diffraction) are located at a distance of

from the center of the pattern. In the formula, m is the order of the minimum,

the wavelenght,

the distance of the screen from the slit and

the width of the slit.
In our problem, the distance of the first-order band (m=1) is

. The distance of the screen is D=86 cm while the wavelength is

. Using these data and re-arranging the formula, we can find a, the width of the slit:
Loudness of a sound wave is directly proportional to the intensity of the sound wave. In other words, when one increases, other also increases and vice-versa
Hope this helps!
5-ohm
Extra
Variable
120-ohm
Variable
Pg. 614
1)
p = 2.4 * 10^5 Pa
T = 18° C + 273.15 = 291.15 k
r = 0.25 m => V = [4/3]π(r^3) = [4/3]π(0.25m)^3 = 0.06545 m^3 = 65.45 L
Use ideal gas equation: pV = nRT => n = pV / RT = [2.4*10^5 Pa * 0.06545 m^3] / [8.31 J/k*mol * 291.15k] = 6.492 mol
Avogadro number = 1 mol = 6.022 * 10^23 atoms
Number of atoms = 6.492 mol * 6.022 *10^23 atom/mol = 39.097 * 10^23 atoms = 3.91 * 10^24 atoms
2) Double atoms => double volume
V2 / V1 = r2 ^3 / r1/3
2 = r2 ^3 / r1 ^3 => r2 ^3 = 2* r1 ^3
r2 = [∛2]r1
The factor is ∛2