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
Answer:
0.82 MPa
Explanation:
the change in pressure 'σ'=160kPa
K= σ/∈ => σ/3∈
K= 160/(3 x 0.065)
K=820 kPA=0.82 MPa
Thus,the bulk modulus of the tissue 'K' is 0.82 MPa
The block is made of A) Tin, as its specific heat capacity is
Explanation:
When an amount of energy Q is supplied to a sample of material of mass m, the temperature of the material increases by , according to the following equation
:
where is the specific heat capacity of the material.
In this problem, we have:
m = 2 kg = 2000 g is the mass of the unknown material
is the amount of energy supplied to the block
is the change in temperature of the material
Solving the equation for , we can find the specific heat capacity of the unknown sample:
And by comparing with tabular values, we can find that this value is approximately the specific heat capacity of tin.
Learn more about specific heat capacity:
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Answer:
A
Explanation:
because u are subtracting if this is from flvs that is what i did and it was right