Answer:

Explanation:
Given that

From the diagram

By differentiating with time t

When x= 10 m

θ = 64.53°
Now by putting the value in equation



Therefore rate of change in the angle is 0.038\ rad/s
weather station - an area where weather data...
satellite - sends pictures...
weather balloon - filled with helium...
radar - sends out signals...
I think the correct answer from the choices listed above is option A. Wave motion is a movement of energy through space or a medium . Some waves are visible light waves, heat waves, sound waves and the like. Hope this answers the question.
Answer:
A. The photographer will get to the jeep before the rhinocerous
Explanation:
Δv = Δd/Δt
we can rearrange for time
Δt = Δd/Δv
For the photographer:
Distance is 10m and moves at 6m/s
Δt = 10m/6m/s
Δt = 1.67s
For the rhinocerous
Distance is 16m and moves at 8m/s
Δt = 16m/8m/s
Δt = 2.00s
The distances were to get to the jeep, the photographer makes it to the jeep in a shorter amount of time than the rhino
The warm air gets pushed up, this also could cause rain, and possibly high winds that can cause a tornado to form