In SI units, its acceleration is (the distance from A to C) / 32 m/s^2 .
The wavelength of the note is

. Since the speed of the wave is the speed of sound,

, the frequency of the note is

Then, we know that the frequency of a vibrating string is related to the tension T of the string and its length L by

where

is the linear mass density of our string.
Using the value of the tension, T=160 N, and the frequency we just found, we can calculate the length of the string, L:
Answer:
3.51s
Explanation:
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A cat falls from a tree (with zero initial velocity) at time t = 0. How far does the cat fall between t = 1/2 and t = 1 s? Use Galileo's formula v(t) = −9.8t m/s.
Answer:
y(1s) - y(1/2s) = - 3.675 m
The cat falls 3.675 m between time 1/2 s and 1 s.
Explanation:
Given data
time=1/2 sec to 1 sec
v(t)=-9.8t m/s
To find
Distance
Solution
As the acceleration as first derivative of velocity with respect to time
So
acceleration(-g)= dv/dt
Solve it
dv = a dt
dv = -g dt
v - v₀ = -gt
v= dy/dt
dy = v dt
dy = ( v₀ - gt ) dt
y(1s) - y(1/2s) = ( v₀ ) ( 1 - 1/2 ) - ( g/2 )[ ( t1)² -( t1/2s )² ]
y(1s) - y(1/2s) = ( - 9.8/2 ) [ ( 1 )² - ( 1/2 )² ]
y1s - y1/2s = ( - 4.9 m/s² ) ( 3/4 s² )
y(1s) - y(1/2s) = - 3.675 m
The cat falls 3.675 m between time 1/2 s and 1 s.