Answer:
These are Diffraction Grating Questions.
Q1. To determine the width of the slit in micrometers (μm), we will need to use the expression for distance along the screen from the center maximum to the nth minimum on one side:
Given as
y = nDλ/w Eqn 1
where
w = width of slit
D = distance to screen
λ = wavelength of light
n = order number
Making x the subject of the formula gives,
w = nDλ/y
Given
y = 0.0149 m
D = 0.555 m
λ = 588 x 10-9 m
and n = 3
w = 6.6x10⁻⁵m
Hence, the width of the slit w, in micrometers (μm) = 66μm
Q2. To determine the linear distance Δx, between the ninth order maximum and the fifth order maximum on the screen
i.e we have to find the difference between distance along the screen (y₉-y₅) = Δx
Recall Eqn 1, y = nDλ/w
given, D = 27cm = 0.27m
λ = 632 x 10-9 m
w = 0.1mm = 1.0x10⁻⁴m
For the 9th order, n = 9,
y₉ = 9 x 0.27 x 632 x 10-9/ 1.0x10⁻⁴m = 0.015m
Similarly, for n = 5,
y₅ = 5x 0.27 x 632 x 10-9/ 1.0x10⁻⁴m = 0.0085m
Recall, Δx = (y₉-y₅) = 0.015 - 0.0085 = 0.0065m
Hence, the linear distance Δx between the ninth order maximum and the fifth order maximum on the screen = 6.5mm
Here's a formula that's simple and useful, and if you're really in
high school physics, I'd be surprised if you haven't see it before.
This one is so simple and useful that I'd suggest memorizing it,
so it's always in your toolbox.
This formula tells how far an object travels in how much time,
when it's accelerating:
Distance = (1/2 acceleration) x (Time²).
D = 1/2 A T²
For your student who dropped an object out of the window,
Distance = 19.6 m
Acceleration = gravity = 9.8 m/s²
D = 1/2 G T²
19.6 = 4.9 T²
Divide each side by 4.9 : 4 = T²
Square root each side: 2 = T
When an object is dropped in Earth gravity,
it takes 2 seconds to fall the first 19.6 meters.
Answer:
It you get a sushi for work, for a party, then it leads to people being happy, as an independant variable
It is changing the molecular structure of a protein
Explanation:
Answer:A) WHICH is 0kgm/s
Explanation:
Answer:
Total mass of combination = 2+3+5 = 10kg.
Acceleration produced = 2m/s^2
hence force =( total mass × acceleration)= (2×10)= 20 N.
Net force on 3kg block = acceleration × mass = (2 × 2 )= 4 N
applied force on 2 kg block = 20N
Force between 2 kg and 3 kg block = (20-4) = 16N. ans
Net force on 3 kg block = 3 × 2 =6N.
Applied force on 3 kg block due to 2 kg block = 16N.
hence, force between 3 kg and 5 kg block = (16-6) = 10N .
answers:-
(a) 20 N
(b) 16N
(c) 10 N
