The answer to this question is amplitude
Answer:
The speed of light is that medium is 281907786.2 m/s.
Explanation:
since the critical angle is Фc = 430, we know that the refractive index is given by:
n = 1/sin(Фc)
= 1/sin(430)
= 1.06
then if n is the refractive index of the medium and c is the speed of light, then the speed of light in the medium is given by:
v = c/n
= (3×10^8)/(1.06)
= 281907786.2 m/s
Therefore, the speed of light is that medium is 281907786.2 m/s.
X Represents the distance the spring is stretched or compressed away from its equilibrium or rest position.
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:
(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