Answer:
h = 599.5 m
Explanation:
Given,
height of structure = 828 m
weight of the tourist = 184 lb
= 184 x 0.45359 = 83.43 Kg
Potential energy = 187000 J
PE = m gd


h = 228.5 m
Height of the room above the ground.
h = 828 - 228.5
h = 599.5 m
Height of the floor above ground is equal to 599.5 m.
Firstly, let's convert the velocities in km/hr to m/s
32*1000/3600=8.89m/s
54*1000/3600=15m/s
From the formula, acceleration=V-U/t
15-8.89/8=0.76m/s²
hope this helps.
D. March because it is just below the 1 million marker on the graph and it is the only one that low.
Clever problem.
We know that the beat frequency is the DIFFERENCE between the frequencies of the two tuning forks. So if Fork-A is 256 Hz and the beat is 6 Hz, then Fork-B has to be EITHER 250 Hz OR 262 Hz. But which one is it ?
Well, loading Fork-B with wax increases its mass and makes it vibrate SLOWER, and when that happens, the beat drops to 5 Hz. That means that when Fork-B slowed down, its frequency got CLOSER to the frequency of Fork-A ... their DIFFERENCE dropped from 6 Hz to 5 Hz.
If slowing down Fork-B pushed it CLOSER to the frequency of Fork-A, then its natural frequency must be ABOVE Fork-A.
The natural frequency of Fork-B, after it gets cleaned up and returns to its normal condition, is 262 Hz. While it was loaded with wax, it was 261 Hz.