Answer:
Waves with high frequencies have shorter wavelengths that work better than low frequency waves for successful echolocation.
Explanation:
To understand why high-frequency waves work better than low frequency waves for successful echolocation, first we have to understand the relation between frequency and wavelength.
The relation between frequency and wavelength is given by
λ = c/f
Where λ is wavelength, c is the speed of light and f is the frequency.
Since the speed of light is constant, the wavelength and frequency are inversely related.
So that means high frequency waves have shorter wavelengths, which is the very reason for the successful echolocation because waves having shorter wavelength are more likely to reach and hit the target and then reflect back to the dolphin to form an image of the object.
Thus, waves with high frequencies have shorter wavelengths that work better than low frequency waves for successful echolocation.
NO musical instrument produces a 'pure' tone with only a
single frequency in it.
EVERY instrument produces more or less harmonics (multiples)
in addition to the basic frequency it's playing.
The percussion instruments (drums etc) are the richest producers
of bunches of different frequencies.
Fuzzy electric guitars are next richest.
The strings and brass instruments are moderate producers of
harmonics ... I can't remember which is greater than the other.
Then come the woodwinds ... clarinet, oboe, etc.
The closest to 'pure' tones of single frequency are the sounds
made by the flute and piccolo, but even these are far from 'pure'.
The only way to get a true single-frequency sound is from an
electronic 'sine wave' generator.
Answer:
α = 2,857 10⁻⁵ ºC⁻¹
Explanation:
The thermal expansion of materials is described by the expression
ΔL = α Lo ΔT
α = 
in the case of the bar the expansion is
ΔL = L_f - L₀
ΔL= 1.002 -1
ΔL = 0.002 m
the temperature variation is
ΔT = 100 - 30
ΔT = 70º C
we calculate
α = 0.002 / 1 70
α = 2,857 10⁻⁵ ºC⁻¹
All of the above.
if your car is on a steep hill it needs force to stop moving and to speed up