The question completely dissolves into nothingness when we recall that the
frequency and wavelength of a wave are inextricably bound to each other.
They don't change independently, and if you know one, then you know the
other. So the "amount of bending" compared to frequency, and the "amount
of bending" compared to wavelength, are necessarily identical.
Yes - the amount of absorption can depend on the frequency.
Frequency
Gene Pool
No.
More frequence = more energy.
The amount of bending certainly CHANGES depending on the frequency, but there is no simple relationship between frequency (or wavelength) on the one hand, and the index of refraction (and therefore the amount of bending) on the other. If by "infer" you mean to calculate this depending on the properties of the material, I don't think there is an obvious way to do this.
Yes - the amount of absorption can depend on the frequency.
The greater the wavelength, the higher the bending will be. Diffraction can be described best through sound, you can hear a loud stereo in different rooms.
The Hotbird network search frequency will depend on where you live and how far you are from a tower.
standard frequency in India is 50Hz
The speed of a wave does NOT depend on its frequency. If it did you would not be able to enjoy an open air concert since the bass sounds would reach you at a different time from the trebles so that the music would be all out of synch.
Frequency
yes, the pitch of sound does, in fact, depend on the frequency of the sound wave.
The speed of a wave depends only on the mechanical or electrical characteristics of the medium or environment through which the wave propagates. It doesn't depend on the wave's frequency or wavelength.
what type of material the pipe is made out of.
Impedance of a coil or a capacitor does depend on the excitation frequency,but resistance has no relationship to frequency.
The speed of a wave doesn't depend on its frequency.REASON:According to the formulaV=frequency * Lambdaso,V/ Lambda= frequencyHere,Frequency is inversely proportional to the wavelength. so, If we increase the value of frequency then by same amount frequency will decrease and will cancel out the effect of each other the "V" will remain constant.