Answer:
2,4,5
Air pressure is created by the weight of the atmosphere pushing on Earth’s surface.
Denser air is heavier than less dense air.
Air is less dense at higher altitudes.
Answer:
![\dfrac{d\theta}{dt}=0.038\ rad/s](https://tex.z-dn.net/?f=%5Cdfrac%7Bd%5Ctheta%7D%7Bdt%7D%3D0.038%5C%20rad%2Fs)
Explanation:
Given that
![\dfrac{dx}{dt}= -1\ m/s](https://tex.z-dn.net/?f=%5Cdfrac%7Bdx%7D%7Bdt%7D%3D%20-1%5C%20m%2Fs)
From the diagram
![tan\theta=\dfrac{21}{x}](https://tex.z-dn.net/?f=tan%5Ctheta%3D%5Cdfrac%7B21%7D%7Bx%7D)
By differentiating with time t
![sec^2\theta \dfrac{d\theta}{dt}=-\dfrac{21}{x^2}\dfrac{dx}{dt}](https://tex.z-dn.net/?f=sec%5E2%5Ctheta%20%5Cdfrac%7Bd%5Ctheta%7D%7Bdt%7D%3D-%5Cdfrac%7B21%7D%7Bx%5E2%7D%5Cdfrac%7Bdx%7D%7Bdt%7D)
When x= 10 m
![tan\theta=\dfrac{21}{10}](https://tex.z-dn.net/?f=tan%5Ctheta%3D%5Cdfrac%7B21%7D%7B10%7D)
θ = 64.53°
Now by putting the value in equation
![sec^2\theta \dfrac{d\theta}{dt}=-\dfrac{21}{x^2}\dfrac{dx}{dt}](https://tex.z-dn.net/?f=sec%5E2%5Ctheta%20%5Cdfrac%7Bd%5Ctheta%7D%7Bdt%7D%3D-%5Cdfrac%7B21%7D%7Bx%5E2%7D%5Cdfrac%7Bdx%7D%7Bdt%7D)
![sec^264.53^{\circ} \dfrac{d\theta}{dt}=-\dfrac{21}{10^2}\times (-1)](https://tex.z-dn.net/?f=sec%5E264.53%5E%7B%5Ccirc%7D%20%5Cdfrac%7Bd%5Ctheta%7D%7Bdt%7D%3D-%5Cdfrac%7B21%7D%7B10%5E2%7D%5Ctimes%20%28-1%29)
![\dfrac{d\theta}{dt}=0.038\ rad/s](https://tex.z-dn.net/?f=%5Cdfrac%7Bd%5Ctheta%7D%7Bdt%7D%3D0.038%5C%20rad%2Fs)
Therefore rate of change in the angle is 0.038\ rad/s