Answer:
T = 4.42 10⁴ N
Explanation:
this is a problem of standing waves, let's start with the open tube, to calculate the wavelength
λ = 4L / n n = 1, 3, 5, ... (2n-1)
How the third resonance is excited
m = 3
L = 192 cm = 1.92 m
λ = 4 1.92 / 3
λ = 2.56 m
As in the resonant processes, the frequency is maintained until you look for the frequency in this tube, with the speed ratio
v = λ f
f = v / λ
f = 343 / 2.56
f = 133.98 Hz
Now he works with the rope, which oscillates in its second mode m = 2 and has a length of L = 37 cm = 0.37 m
The expression for standing waves on a string is
λ = 2L / n
λ = 2 0.37 / 2
λ = 0.37 m
The speed of the wave is
v = λ f
As we have some resonance processes between the string and the tube the frequency is the same
v = 0.37 133.98
v = 49.57 m / s
Let's use the relationship of the speed of the wave with the properties of the string
v = √ T /μ
T = v² μ
T = 49.57² 18
T = 4.42 10⁴ N
Answer:
P = m g v
Explanation:
Power is defined as the relationship between work and time
P = W / t
The work is defined by the scalar product of the force and the displacement, in the case of a drop falling the vertical force and the displacement has the same direction, therefore the angle between them is zero and the cosines of zero is one (cos 0 = 1)
W = F y
the gravitational force is proportional to the weight of the drop (F = mg)
W = mg y
we substitute
P = m g y / t
the terminal velocity of the drop is constant
P = m g v
Hope this helps :) If in need of clarification, feel free to ask!
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