Answer:
The fundamental resonance frequency is 172 Hz.
Explanation:
Given;
velocity of sound, v = 344 m/s
total length of tube, Lt = 1 m = 100 cm
height of water, hw = 50 cm
length of air column, L = Lt - hw = 100 cm - 50 cm = 50 cm
For a tube open at the top (closed pipe), the fundamental wavelength is given as;
Node to anti-node (N ---- A) : L = λ / 4
λ = 4L
λ = 4 (50 cm)
λ = 200 cm = 2 m
The fundamental resonance frequency is given by;
Therefore, the fundamental resonance frequency is 172 Hz.
Answer:
0.00288 J
Explanation:
We know that
W= Fds
F = −α∆s^4
α = 45 N/m^4 and ∆s = displacement
W= −α∆s^4ds
integrating both the sides from s= 0 to 0.2
W= 45/5×0.2^5= 0.00288 J
Answer:
20 Hz
15.8 times
Explanation:
A
Although the range of frequency for any human's ear is usually said to be between 20 Hz and 20 kHz. And since the question asked for the least intense frequency, that has to be 20 Hz. Essentially the frequency most people perceive the least intense sound is 20 Hz.
B
A 100-Hz sound must be 10^1.2 times or 15.8 times more intense compared to a 1000-Hz sound to be perceived as equal to 60 phons of loudness
The combination of all of the forces acting on object is net force.
Answer:
Acceleration = 5.77 m/s²
Distance cover in 13 seconds = 487.56 meter
Explanation:
Given:
Final velocity of mobile device = 75 m/s
initial velocity of mobile device = 0 m/s
Time taken = 13 seconds
Find:
Acceleration
Distance cover in 13 seconds
Computation:
v = u + at
75 = 0 + (a)(13)
13a = 75
a = 5.77
Acceleration = 5.77 m/s²
s = ut + (1/2)(a)(t²)
s = (0)(t) + (1/2)(5.77)(13²)
Distance cover in 13 seconds = 487.56 meter