Answer:
1.16cm were cut off the end of the second pipe
Explanation:
The fundamental frequency in the first pipe is,
<em><u>Since the speed of sound is not given in the question, we would assume it to be 340m/s</u></em>
f1 = v/4L, where v is the speed of sound and L is the length of the pipe
266 = 340/4L
L = 0.31954 m = 0.32 m
It is given that the second pipe is identical to the first pipe by cutting off a portion of the open end. So, consider L’ be the length that was cut from the first pipe.
<u>So, the length of the second pipe is L – L’</u>
Then, the fundamental frequency in the second pipe is
f2 = v/4(L - L’)
<u>The beat frequency due to the fundamental frequencies of the first and second pipe is</u>
f2 – f1 = 10hz
[v/4(L - L’)] – 266 = 10
[v/4(L – L’)] = 10 + 266
[v/4(L – L’)] = 276
(L - L’) = v/(4 x 276)
(L – L’) = 340/(4 x 276)
(L – L’) = 0.30797
L’ = 0.31954 – 0.30797
L’ = 0.01157 m = 1.157 cm ≅ 1.16cm
Hence, 1.16 cm were cut from the end of the second pipe
Answer: The acceleration of the object is 0.67m/s^2 west.
Explanation: Here we are given the initial velocity and final velocity as well as the time taken. Acceleration is the change in velocity per unit time, thus the equation becomes.
a=dv/t
a=vf-vi/t
a=-2.1-4.7/3.9
a= 0.66m/s^2 west
Hi There,
This is False.
Hope this helped!
D. similar types of rocks and fossils