Answer:
a) 2.4 mm
b) 1.2 mm
c) 1.2 mm
Explanation:
To find the widths of the maxima you use the diffraction condition for destructive interference, given by the following formula:
a: width of the slit
λ: wavelength
m: order of the minimum
for little angles you have:
y: height of the mth minimum
a) the width of the central maximum is 2*y for m=1:
b) the width of first maximum is y2-y1:
![w=y_2-y_1=\frac{(500*10^{-9}m)(1.2m)}{0.50*10^{-3}m}[2-1]=1.2mm](https://tex.z-dn.net/?f=w%3Dy_2-y_1%3D%5Cfrac%7B%28500%2A10%5E%7B-9%7Dm%29%281.2m%29%7D%7B0.50%2A10%5E%7B-3%7Dm%7D%5B2-1%5D%3D1.2mm)
c) and for the second maximum:
![w=y_3-y_2=\frac{(500*10^{-9}m)(1.2m)}{0.50*10^{-3}m}[3-2]=1.2mm](https://tex.z-dn.net/?f=w%3Dy_3-y_2%3D%5Cfrac%7B%28500%2A10%5E%7B-9%7Dm%29%281.2m%29%7D%7B0.50%2A10%5E%7B-3%7Dm%7D%5B3-2%5D%3D1.2mm)
Initial speed(u)=0m/s
Final speed(v)= 27m/s
Time(t)=7.6s
Use the equation of motion: v = u + at
27 = 0 + a(7.6)
27/7.6 = a
a = 3.55 m/s^2 (3 s.f)
D, I believe is the correct answer. Hope I answered your question, have a good day.<span />
STEP ONE:
Let you and your friend stand as far away as possible from a large reflecting wall and clap your hands rapidly at a regular rate.
STEP TWO:
Adjust this rate until each clap just coincides with the return of an echo of its predecessor, or until clap and echo are heard as equally spaced.
STEP THREE:
Use a stopwatch to find the time between claps, t. Make a rough measurement of distance to the wall, s. Thus the speed of sound, v = 2s/t
Rate of change of momentum = impact force
(m*v-m*u)/t = F
4000*20/t = 80000 (note: v is zero as it stopped)
<span>soo, t = 1 sec</span>
