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
In the given question, one important information for getting to the actual solution is not given and that is the atmospheric pressure. To find the approximate absolute pressure, it is needed to add the value of atmospheric pressure with the gage pressure.
Atmospheric pressure = 100 kPa
Then
Absolute pressure = 156 + 100 kPa
= 256 KPa.
Let <em>v</em> be the object's volume. The object displaces 0.4<em>v</em> cm³ of water, which, at a density of about 0.997 g/cm³, has a weight of
<em>b</em> = (0.000997 kg/cm³) (0.4 <em>v</em> cm³) <em>g</em> ≈ 0.000391<em>v</em> N
(and <em>b</em> is also the magnitude of the buoyant force). Then the net force on the object while it's floating in water is
∑ <em>F</em> = <em>b</em> - <em>mg</em> = 0
so that <em>b</em> = <em>mg</em>, where <em>mg</em> is the object's weight. This weight never changes, so the object feels the same buoyant force in each liquid.
(a) In methanol, we have
<em>b</em> = 0.000391<em>v</em> N = (0.00079 kg/cm³) (<em>pv</em> cm³) <em>g</em>
where <em>p</em> is the fraction of the object's volume that is submerged. Solving for <em>p</em> gives
<em>p</em> = (0.000391 N) / ((0.00079 kg/cm³) <em>g</em>) ≈ 0.0505 ≈ 5.05%
(b) In carbon tetrachloride, we have
<em>b</em> = 0.000391<em>v</em> N = (0.00158 kg/cm³) (<em>pv</em> cm³) <em>g</em>
==> <em>p</em> ≈ 0.0253 ≈ 2.53%
Explanation:
The relation between angular frequency and frequency is directly proportional:
So, if the angular frequency decreases, the frequency also decreases.
The relation between angular frequency and period is inversely proportional:
So, if the angular frequency decreases, the period increases.
Answer:
1.59 Tesla per second
Explanation:
Given that
Radius of the conductor loop, r = 1 m,
Maximum induced emf to the loop, e = 5 V,
We take an assumption, we assume that the rate which the magnetic field is changing is dB / dt.
We then go ahead to apply Farady's law of electromagnetic induction
e = rate of change of magnetic flux
e = dФ /dt
e = A * dB / dt, on substituting we see that
A = πr²
A = π * 1²
A = π, plug in the value for A
5 = 3.14 * dB / dt
dB / dt = 5/3.142
dB / dt = 1.59 Tesla per second
