Answer:
ax = -3.29[m/s²]
ay = -1.9[m/s²]
Explanation:
We must remember that acceleration is a vector and therefore has magnitude and direction.
In this case, it is accelerating downwards, therefore for a greater understanding we will make a diagram of said vector, this diagram is attached.
![a_{x}=-3.8*cos(30) = -3.29 [m/s^{2}]\\ a_{y}=-3.8*sin(30) = -1.9 [m/s^{2}]](https://tex.z-dn.net/?f=a_%7Bx%7D%3D-3.8%2Acos%2830%29%20%3D%20-3.29%20%5Bm%2Fs%5E%7B2%7D%5D%5C%5C%20a_%7By%7D%3D-3.8%2Asin%2830%29%20%3D%20-1.9%20%5Bm%2Fs%5E%7B2%7D%5D)
Answer:
The tension is 
Explanation:
The free body diagram of the question is shown on the first uploaded image From the question we are told that
The distance between the two poles is 
The mass tied between the two cloth line is 
The distance it sags is 
The objective of this solution is to obtain the magnitude of the tension on the ends of the clothesline
Now the sum of the forces on the y-axis is zero assuming that the whole system is at equilibrium
And this can be mathematically represented as

To obtain
we apply SOHCAHTOH Rule
So 
![\theta = tan^{-1} [\frac{opp}{adj} ]](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20tan%5E%7B-1%7D%20%5B%5Cfrac%7Bopp%7D%7Badj%7D%20%5D)
![= tan^{-1} [\frac{1}{7}]](https://tex.z-dn.net/?f=%3D%20tan%5E%7B-1%7D%20%5B%5Cfrac%7B1%7D%7B7%7D%5D)






Answer:
Art
Explanation:
Polly's line is linear, while arts line is going up with constant velocity. There for art is going faster.
Answer:
The index of refraction of the liquid is 1.35.
Explanation:
It is given that,
Critical angle for a certain air-liquid surface, 
Let n₁ is the refractive index of liquid and n₂ is the refractive index of air, n₂ =1
Using Snell's law for air liquid interface as :




So, the index of refraction of the liquid is 1.35. Hence, this is the required solution.