A flutist assembles her flute in a room where the speed of sound is 340 m/s. When she plays the note A, it is in perfect tune with a 440 Hz tuning fork. After a few minutes, the air inside her flute has warmed to where the speed of sound is 347 m/s .

Explanation:

Let L = length of flute used in playing A-note (open ends forming antinodes, L= lamda/2)

f1 = v1/2L = 440

340/2L = 440

2L = 340/440

= 0.386 m.

At v2 = 347 m/s,

f2 = v2/2L

= 347/2L

= 347 * 440/340

= 449.06 Hz

Beat frequency, f = f2 - f1

= 347 * 440/340 - 440

f = 9.06 beats/s

B.

f2 > f1 because now length is not able to keep ratio with increased speed of sound, so lets increase L to (L+x), f2' = f1 (again resonates) or beats f = 0

347/(2*(L + x)) = 440

2L + 2x = 347/440

340/440 + 2x = 347/440

2x = 7/440

x = 0.00795 m

= 7.95 mm.

in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.

d is the correct answer