The frequency of note C3 is 131 
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<u>Explanation:</u>
Frequency is the measure of repetition of same thing a certain number of times. So frequency is inversely proportional to the wavelength. As wavelength is distance between two successive crests or troughs in a sound wave.
And frequency is the completion of number of cycles in a given time in sound waves. The frequency and wavelength are inversely proportional to each other with velocity of sound being the proportionality constant.
Thus, here the speed of sound is given as 343 m/s, the wavelength of the note is also given as 2.62 m, then frequency will be as follows:
Thus,
So the frequency of note C3 is 131 .
Answer:
power=work done÷time taken
2×5=10
10÷10=1
ans 1J per second
Answer: object B is negatively charged, object C is positively charged and object D is also positively charged
Explanation: since unlike charges attract and like charges repel, for object A which is positively charged and B to attract B must be negatively charged and then for B which is negatively charged and C to attract C must be positively charged and for C and D to repel they have to be of thesame charge which means D is positive as well.
Thank you for your question, what you say is true, the gravitational force exerted by the Earth on the Moon has to be equal to the centripetal force.
An interesting application of this principle is that it allows you to determine a relation between the period of an orbit and its size. Let us assume for simplicity the Moon's orbit as circular (it is not, but this is a good approximation for our purposes).
The gravitational acceleration that the Moon experience due to the gravitational attraction from the Earth is given by:
ag=G(MEarth+MMoon)/r2
Where G is the gravitational constant, M stands for mass, and r is the radius of the orbit. The centripetal acceleration is given by:
acentr=(4 pi2 r)/T2
Where T is the period. Since the two accelerations have to be equal, we obtain:
(4 pi2 r) /T2=G(MEarth+MMoon)/r2
Which implies:
r3/T2=G(MEarth+MMoon)/4 pi2=const.
This is the so-called third Kepler law, that states that the cube of the radius of the orbit is proportional to the square of the period.
This has interesting applications. In the Solar System, for example, if you know the period and the radius of one planet orbit, by knowing another planet's period you can determine its orbit radius. I hope that this answers your question.
