Answer: 18.27°
Explanation:
Given
Index of refraction of blue light, n(b) = 1.64
Wavelength of blue light, λ(b) = 440 nm
Index of refraction of red light, n(r) = 1.595
Wavelength of red light, λ(r) = 670 nm
Angle of incident, θ = 30°
Angle of refraction of red light is
θ(r) = sin^-1 [(n(a)* sin θ) / n(r)], where n(a) = index of refraction of air = 1
So that,
θ(r) = sin^-1 [(1 * sin 30) / 1.595]
θ(r) = sin^-1 (0.5 / 1.595)
θ(r) = sin^-1 0.3135
θ(r) = 18.27°
Answer:
v_max = (1/6)e^-1 a
Explanation:
You have the following equation for the instantaneous speed of a particle:
(1)
To find the expression for the maximum speed in terms of the acceleration "a", you first derivative v(t) respect to time t:
![\frac{dv(t)}{dt}=\frac{d}{dt}[ate^{-6t}]=a[(1)e^{-6t}+t(e^{-6t}(-6))]](https://tex.z-dn.net/?f=%5Cfrac%7Bdv%28t%29%7D%7Bdt%7D%3D%5Cfrac%7Bd%7D%7Bdt%7D%5Bate%5E%7B-6t%7D%5D%3Da%5B%281%29e%5E%7B-6t%7D%2Bt%28e%5E%7B-6t%7D%28-6%29%29%5D)
(2)
where you have use the derivative of a product.
Next, you equal the expression (2) to zero in order to calculate t:
![a[(1)e^{-6t}-6te^{-6t}]=0\\\\1-6t=0\\\\t=\frac{1}{6}](https://tex.z-dn.net/?f=a%5B%281%29e%5E%7B-6t%7D-6te%5E%7B-6t%7D%5D%3D0%5C%5C%5C%5C1-6t%3D0%5C%5C%5C%5Ct%3D%5Cfrac%7B1%7D%7B6%7D)
For t = 1/6 you obtain the maximum speed.
Then, you replace that value of t in the expression (1):
hence, the maximum speed is v_max = ((1/6)e^-1)a
(1) Speed is the ratio of the total distance covered by the object and the total time it takes for him to finish it.
Speed = distance / time
In this item, we are given that the distance is 20 kilometers and that the time it takes for the trip is 2 hours. Substituting the known values,
Speed = 20 kilometers / 2 hours
speed = 20 km/h
(2) Velocity on the other hand takes into account the displacement of the object from his original position. It is assumed that Jeremie was basically back to his original position after two hours. Hence, the velocity is equal to zero.
