When air is forced into the open pipe,
L = 
where n represents any whole number such as 1,2,3,4, etc., and λ denotes the wavelength of the oscillation
This implies λ=
It is important to note that n=1 corresponds to the fundamental frequency, n=2 corresponds to the first harmonic, and so forth.
Thus, the third harmonic will be for n=4
With L=6m and n=4, solving for λ yields:
λ=
=3m
The connection between frequency (f), sound speed (c), and wavelength (λ) is given by:
c=f.λ or f= 
Therefore, f=
≈115 Hz
<span>You are presented with a circuit that includes a 6.0-v battery, a 4.0-ohm resistor, a 0.60 microfarad capacitor, an ammeter, and a switch all connected in series. Your task is to determine the current reading once the switch is closed. Ohm's law should be used, which states V = IR where V signifies voltage, I indicates current, and R represents resistance.</span>
V = IR
I = V/R
I = 6 volts / 4 ohms
I = 1.5A
Upon closing the switch, the cathode side plate starts accumulating electrons if it was previously empty. As this process continues, the current diminishes. Eventually, when the capacitor reaches its maximum electron retention, the current will cease. An increased capacitance means a greater capacity for electron storage.
The height is h = 17 10⁶ meters above the surface of Mars. To determine this, we apply Newton's second law according to the universal law of gravitation, represented by F = m a. The centripetal acceleration a is expressed as v² / r. Applying the gravitational force we have G m M / r² = m v² / r. Given that the speed of the object remains constant, we derive v from d / t, where d is the circumference and t is the orbital period. Substituting gives us d = 2π r and v = 2π r / T. Replacing these values leads to the equation G M / r² = (4π² r² / T) / r, so r³ = G M T² / 4π². Converting time into SI units, T = 24.66 h converts to 88776 seconds. Ultimately, the computed value of r is 2,045 10⁶ m, and after subtracting Mars’ radius of 3.39 10⁶ m, we find the height h to be 17 10⁶ m.
