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:
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Explanation:
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D. velocity
Velocity depends on speed and direction
The magnitude of the resultant force is given by the centripetal force, since the car is under a circular motion. So, we have:
The centripetal acceleration is given by:
Where v is the linear speed and r the radius of the circular motion. Replacing this and solving:
a)
• P = F/A
P = pressure = 630 N/m^2
F = force
A = area
F = mg = 0.50 kg x 9.8 m/s^2 = 4.9 N
m= mass
g= gravity
P = F/A
A = F/P
A = 4.9 N / 630 N/m^2 = 7.778 x 10^-3 m^2
b)
• Area of a circle = pi* radius ^2
7.778 x 10^-3 m^2 = pi* radius ^2
√(7.778 x 10^-3 m^2 / pi ) = radius
radius = 0.04976 m
Answers:
a ) 7.778 x 10^-3 m^2
b) 0.04976 m
