Answer:
1.696 nm
Explanation:
For a diffraction grating, dsinθ = mλ where d = number of lines per metre of grating = 5510 lines per cm = 551000 lines per metre and λ = wavelength of light = 467 nm = 467 × 10⁻⁹ m. For a principal maximum, m = 1. So,
dsinθ = mλ = (1)λ = λ
dsinθ = λ
sinθ = λ/d.
Also tanθ = w/D where w = distance of center of screen to principal maximum and D = distance of grating to screen = 1.03 m
From trig ratios 1 + cot²θ = cosec²θ
1 + (1/tan²θ) = 1/(sin²θ)
substituting the values of sinθ and tanθ we have
1 + (D/w)² = (d/λ)²
(D/w)² = (d/λ)² - 1
(w/D)² = 1/[(d/λ)² - 1]
(w/D) = 1/√[(d/λ)² - 1]
w = D/√[(d/λ)² - 1] = 1.03 m/√[(551000/467 × 10⁻⁹ )² - 1] = 1.03 m/√[(1179.87 × 10⁹ )² - 1] = 1.03 m/1179.87 × 10⁹ = 0.000848 × 10⁻⁹ = 0.848 × 10⁻¹² m = 0.848 nm.
w is also the distance from the center to the other principal maximum on the other side.
So for both principal maxima to be on the screen, its minimum width must be 2w = 2 × 0.848 nm = 1.696 nm
So, the minimum width of the screen must be 1.696 nm
For this we use general equation for gases. Our variables represent:
p- pressure
v-volume
t- temperature
P1V1/T1 = P2V2/T2
in this equation we know:
P1,V1 and T1, T2 and V2.
We have one equation and 1 unknown variable.
P2 = T2P1V1/T1V2 = 1.1atm
Answer:
T=75N
Explanation:
In order to solve this problem, we must first start by drawing a free body diagram (see attached picture).
Once we got the free body diagram, we can do an addition of forces so we get:
it equals zero because it's at equilibrium.
So the sum of forces in this case will be:
so now we solve for the tension T:
we need to figure out what W is equal to. But in order to do that, we need to figure out what the volume of the glass is. We know the buoyant force is calculated by multiplying the volume displaces by the density of water and the acceleration of gravity, so we get:
So the volume of the glass will be:
we know the density of water is
so our volume is:
Now we can calculate the weght of the piece of glass, so we get:
W=125N
and now we can calculate the tension:
T=125N-50N
T=75N
A force is described by its MAGNITUDE and by the direction in which it acts