It is an example of Hypermobility.
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
Answer:
She is likely to crash because her flight gradient is lesser than the flight gradient required gradient to avoid crashing
Explanation:
The given parameters are;
The required gradient of the plane Ashley is flying needs to reach in order to take off and not crash = 360 m/km
The initial elevation of the plane Ashley is flying = Sea level = 0 m
The goal Ashley intends to make = Elevation of 1000 m at 2.8 km. distance
∴ Ashley's goal = Traveling from sea level to 1000 m at 2.8 km horizontal distance
We have;
The gradient = Rate of change of elevation/(Horizontal distance)
Therefore;
The gradient of Ashley's flight = (1000 - 0)/(2.8 - 0) = 357.143 m/km
The gradient of Ashley's flight ≈ 357.143 m/km which is lesser than the required 360 m/km in order to take off and not crash, therefore, she will crash.
Answer: a) 16Hz, 3m b) 48Hz, 1mc) 80Hz, 0.6m
Explanation:
a) Fundamental frequency in string is represented as Fo = V/2L where;
Fo is the fundamental frequency
V is the speed of the transverse wave = 48m/s
L is the length of the wire. = 1.50m
Substituting this values in the formula given we have;
Fo = 48/2(1.5)
Fo = 48/3
Fo = 16Hz
The fundamental tone is therefore 16Hz
Using v =f¶
Where f is the frequency and ¶ is the wavelength, the wavelength of the fundamental note will be;
¶ = v/fo
¶ = 48/16 = 3m
b) Overtones or harmonics is the multiple integral of the fundamental frequency. The multiples are I'm arithmetical progression.
First overtone f1 = 2fo
Second overtone f2 = 3fo etc.
Since fo = 16Hz
Second overtone f2 = 3×16 = 48Hz
¶ = v/f2 = 48/48
¶ = 1m
c) Fourth harmonic or overtone will be f4 = 5fo
F4 = 5×16 = 80Hz
The fourth harmonic is therefore 80Hz
¶ = v/f4 = 48/80
¶ = 0.6m
No, because terminal velocity is when the acceleration of the Earth’s gravity is balanced by the air resistance of the atmosphere.
