When Sequoia measured the sound from the tube open at one end and closed at the other, she only observed three, equally-spaced, peaks in the Fourier transform window, and thus recorded three frequencies (150 Hz, 450 Hz, and 750 Hz). If the length of the tube was 0.55 m, what would be the speed of sound that she would calculate using the fundamental frequency she measured?

v= 330m/s

Explanation: the length (l) of an open closed pipe is related to the wavelength (λ) as shown in the formulae below

l = λ/4

where l = 0.55m, hence we have that

0.55 = λ/4

λ = 0.55*4

λ = 2.2m

from the question, it can be seen that the fundamental frequency is 150Hz.

Recall that we have v=fλ, where v = speed of wave, f= frequency, λ= wavelength

v =2.2 * 150

v = 330 m/s

5 w for 1hr = 5x60x60 joules=300x60=3000x6=18,000j as total energy conversion. efficiency of bulb needed for radiant to thermal within 18000

305.7

explanation:

density =m/v

density =917/3.00

density = 305.

approximately = 305.7

in four significant figures