The wavelength of the third resonance of the closed organ pipe is equal to the ratio between the speed of sound and the frequency of the 3rd harmonic:

The relationship between length of a closed pipe and wavelength of the standing waves inside is:

where n is the number of the harmonic. In this case, n=3, so the length of the pipe is

^^^^^^^^^^^^^^^^^^^^^^^^^^^ is correct
The part of a river that would have animals with muscular bodies and adaptations that let survive in turbulent water is in the transition zone, the mid-transition zone to be precise.
Water at the source zone possesses a lot of potential energy and as it flows from the upper reaches the potential energy is turned into kinetic energy when the course of the river begins to gradually level out and this translates into increase in velocity. By the time river water reaches the middle of the transition zone, most of the potential energy would have been turned into kinetic energy and thus water velocity would be quite high here.
Animals living here would develop muscles because of constantly fighting against the strong current to avoid being swept downstream.
Medicine to a patient. That should be calculated based on weight, strength/dosage and possibly other factors
Given:
Gasoline pumping rate, R = 5.64 x 10⁻² kg/s
Density of gasoline, D = 735 kg/m³
Radius of fuel line, r = 3.43 x 10⁻³ m
Calculate the cross sectional area of the fuel line.
A = πr² = π(3.43 x 10⁻³ m)² = 3.6961 x 10⁻⁵ m²
Let v = speed of pumping the gasoline, m/s
Then the mass flow rate is
M = AvD = (3.6961 x 10⁻⁵ m²)*(v m/s)*(735 kg/m³) = 0.027166v kg/s
The gasoline pumping rate is given as 5.64 x 10⁻² kg/s, therefore
0.027166v = 0.0564
v = 2.076 m/s
Answer: 2.076 m/s
The gasoline moves through the fuel line at 2.076 m/s.