The concepts necessary to solve this problem are framed in the expression of string vibration frequency as well as the expression of the number of beats per second conditioned at two frequencies.
Mathematically, the frequency of the vibration of a string can be expressed as
Where,
L = Vibrating length string
T = Tension in the string
Linear mass density
At the same time we have the expression for the number of beats described as
Where
= First frequency
= Second frequency
From the previously given data we can directly observe that the frequency is directly proportional to the root of the mechanical Tension:
If we analyze carefully we can realize that when there is an increase in the frequency ratio on the tight string it increases. Therefore, the beats will be constituted under two waves; one from the first string and the second as a residue of the tight wave, as well
Replacing for n and 202Hz for
The frequency of the tightened is 205Hz
Answer : When a person applies force and covers distance then the person is doing work on an object.
The answer is up . tylrhscjwizn
So for a minute lets ignore the 880 km/h. If it took 4 hours and she flew at 600 km/h 600*4=2400. Now lets Look at the 880 bit. If it took 4 hours and she where to fly at 880 it would've been 880*4=3520. Lets do 2400-600=1800, now we've got the 600 kmh bit done. Now lets see if you fly 880 km/h for one hour then you add 1800+880=2680.
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