Atunnel under a river is 1.30 km long. (a) at what frequencies can the air in the tunnel resonate? (use n to represent the harmonics and n = 1, 2, 3,

f = n * 43/325 Hz, where n = 1, 2, 3, ... Modeling the tunnel as a cylinder that's open at each end, the 1st harmonic will have it's node (minimum motion) at the center of the tunnel and the anti-noes at each end of the tunnel. So the wavelength of the 1st harmonic will be 2 * 1.30 km = 2.60 km. The formula for the frequencies is: f = nv/(2L) where f = frequency n = harmonic number v = speed of sound L = length of tube There is a more accurate formula where L is replaced with (L + 0.8d) where d is the diameter of the tube, but since the diameter of the tunnel isn't given, we'll use the simpler, but less accurate formula. The speed of sound at sea level is approximately 344 m/s, so substituting known values into the formula gives: f = nv/(2L) f = n344m/s / (2*1300 m) f = n*43/325 s Plugging the first few values of n into the formula gives n = 1: 0.132 Hz n = 2: 0.265 Hz n = 3: 0.397 Hz

Magic man is here, but unfortunately he is broke won’t you him out

29. because you will divide each number of months by 12 to find the number of years. which would mean, 5, 12.5, 16, 45, 50, 72.5. which is crazy. anyhow, i think you find the median in the middle. and yea hope your test goes good