Acceleration = vf-vi /t
10-22/3=2.6m/s^2
Answer:
film is at distance of 3.07 cm from lens
Explanation:
Given data
focal length = 3.06 cm
distance = 10.4 m = 1040 cm
to find out
How far must the lens
solution
we apply here lens formula that is
1/f = 1/p + 1/q
here f = 3.06 and p = 1040 so we find q
1/f = 1/p + 1/q
1/3.06 = 1/1040 + 1/q
1/ q = 0.3258
q = 3.0690 cm
so film is at distance of 3.07 cm from lens
f = frequency of the sound wave = 680 hertz
λ = wavelength of the sound wave = 0.5 meters
v = speed of sound wave
we know that , speed of sound wave is given as
speed of sound wave = frequency of sound wave x wavelength of sound wave
v = f λ
inserting the above values in the formula above
v = (680 hertz) (0.5 meters)
v = 340 meter/second
hence the speed of sound wave comes out to be 340 meter/second
