I’m gonna have to say “Ocean waves” as the answer
Part A. For this part, we use two equations for linear
motion:
<span>y = y0 + v0 t + 0.5 g t^2 --->
1</span>
<span>vf = v0 + g t --->
2</span>
First we solve for t using equation 1: y0 = 0 (initial
point at top), y = 250 m, v0 = 0 (at rest)
250 = 0.5 (9.8) t^2
t = 7.143 s
Now we solve for final velocity vf using equation 2:
vf = g t
vf = 9.8 (7.143)
vf = 70 m/s
Part B. First we solve for the time it takes for the sound
to reach the tourist.
t(sound) = 250 / 335 = 0.746 s
Therefore the total time would be:
t = 0.746 s + 0.300 s
t = 1.05 s
<span>Hence there is enough time for the tourist to get out
before the boulder hits him.</span>
Complete Question
The spaceship Intergalactica lands on the surface of the uninhabited Pink Planet, which orbits a rather average star in the distant Garbanzo Galaxy. A scouting party sets out to explore. The party's leader–a physicist, naturally–immediately makes a determination of the acceleration due to gravity on the Pink Planet's surface by means of a simple pendulum of length 1.08m. She sets the pendulum swinging, and her collaborators carefully count 101 complete cycles of oscillation during 2.00×102 s. What is the result? acceleration due to gravity:acceleration due to gravity: m/s2
Answer:
The acceleration due to gravity is
Explanation:
From the question we are told that
The length of the simple pendulum is 
The number of cycles is 
The time take is
Generally the period of this oscillation is mathematically evaluated as

substituting values


The period of this oscillation is mathematically represented as

making g the subject of the formula we have
![g = \frac{L}{[\frac{T}{2 \pi } ]^2 }](https://tex.z-dn.net/?f=g%20%3D%20%5Cfrac%7BL%7D%7B%5B%5Cfrac%7BT%7D%7B2%20%5Cpi%20%7D%20%5D%5E2%20%7D)

Substituting values

Answer: burning of the incense and air drift in the room
Explanation: the smoke will automatically rise to the top of the room and then with air drift if would help spread further around the room and spread the smell
At the same speed as the other ice cube