Answer:
Δd = 7.22 10⁻² m
Explanation:
For this exercise we must use the dispersion relationship of a diffraction grating
d sin θ = m λ
let's use trigonometry
tan θ = y / L
how the angles are small
tant θ = sinθ /cos θ = sin θ
we substitute
sin θ = y / L
d y / L = m λ
y = m λ L / d
let's use direct ruler rule to find the distance between two slits
If there are 500 lines in 1 me, what distance is there between two lines
d = 2/500
d = 0.004 me = 4 10⁻⁶ m
diffraction gratings are built so that most of the energy is in the first order of diffraction m = 1
let's calculate for each wavelength
λ = 656 nm = 656 10⁻⁹ m
d₁ = 1 656 10⁻⁹ 1.7 / 4 10⁻⁶
d₁ = 2.788 10⁻¹ m
λ = 486 nm = 486 10⁻⁹ m
d₂ = 1 486 10⁻⁹ 1.7 / 4 10⁻⁶
d₂ = 2.066 10⁻¹ m
the distance between the two lines is
Δd = d1 -d2
Δd = (2,788 - 2,066) 10⁻¹
Δd = 7.22 10⁻² m
Your hand is not capable of moving fast enough for sound to come out of it.
According to snells law
<span>n1 sin theta1 = n2 sin theta2
</span>n1 = 1.333 (water)
<span>n2 = 2.42 (diamond)</span>
it is given that theta =30 degrees so
by putting the values we have
<span>1.333 sin theta = 2.42 sin 30 </span>
<span>sin theta = (2.42/1.333) *0.5 =65.2 degree
</span>so our conclusion is
<span>the ray's angle of incidence θ1 on the diamond</span> = 65.2 degree.
hope this helps
To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.
From the perspective of angular movement, we find the relationship with the tangential movement of velocity through

Where,
Angular velocity
v = Lineal Velocity
R = Radius
At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is

Where
Angular acceleration
Angular velocity
t = Time
Our values are




Replacing at the previous equation we have that the angular velocity is



Therefore the angular speed of a point on the outer edge of the tires is 66.67rad/s
At the same time the angular acceleration would be



Therefore the angular acceleration of a point on the outer edge of the tires is 