Answer:
6 km/h
Explanation:
V avg = ∆x/∆t = 3km / 30 min ×(60min/1h) = 3 km× 2 /h = 6 km/h
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.
Answer:
4.08 s
Explanation:
Let the passenger took "t" time to catch the train
so in this case the total distance moved by the train + 5 m = total distance moved by the passenger
so we will have
distance moved by train is given as
also the distance moved by passenger
so we will have
t = 4.08 s
F(of spring)=230x=ma=3.5(5)=17.5=230x; x=0.07m.
