A star is located 5.9 light years from Earth.
We know that : 1 light year = 9.46 trillion kilometers.
We will calculate the distance in trillion kilometers multiplying the number of light years by 9.46:
5.9 * 9.46 = 55.814
Answer: The distance is 55.814 trillion km.
<h2>Answers:</h2><h2 /><h2>a) Arrow B</h2><h2>b) Arrow E</h2>
Explanation:
Refraction is a phenomenon in which a wave (the light in this case) bends or changes its direction <u>when passing through a medium with a refractive index different from the other medium.</u> Where the Refractive index is a number that describes how fast light propagates through a medium or material.
According to this, if we observe the rays A an D passing throgh the biconcave lens, we will have two mediums:
1) The air
2)The material of the biconcave lens
This two mediums have different refractive indexes, hence the rays will change the direction.
-For the incident ray A, the corresponding refractive ray is B, because is the ray that bends after passing throgh the lens
-For the incident ray D, the refracted ray is E following the same principle.
U can do what is the strongest paper towel
120 acceleration oh yeah i am sure about this
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
