Answer:
Frequency of the tuning fork[second resonance] = 512 Hz (Approx.)
Explanation:
Given:
Speed of sound = 340 m/s
Length of resonance position above a closed air column = 49.8 cm = 0.498 m
Find:
Frequency of the tuning fork
Computation:
Frequency of the tuning fork[second resonance] = 3v / 4l
Frequency of the tuning fork[second resonance] = 3(340) / 4(0.498)
Frequency of the tuning fork[second resonance] = 512.04
The frequency is 512 Hz.
speed, v = 340 m/s
length, L = 49.8 cm = 0.498 m
let the frequency is f.
1/2mv²=0
1/2(4kg)(v²)=0
2=-v²
square root -2=v
v=1.414
When we push the box from the bottom of the incline towards the top then by work energy theorem we can say that
Work done by all the forces = change in kinetic energy of the system
here we know that
also we know that the length of the incline is given as
now we have
so we have
Answer: idk
Explanation: hahahaha