The speed at which sound travels through the gas in the tube is 719.94m/s
<u>Explanation:</u>
Given:
Frequency, f = 11999Hz
Wavelength, λ = 0.03m
Velocity, v = ?
Sound speed in the tube is calculated by multiplying the frequency v by the wavelength λ.
As the sound loudness changed from a maximum to a minimum, then we know the sound interference in the case changed from constructive interference (the two sound waves are in phase, i.e. peaks are in a line with peaks and so the troughs), to a destructive interference (peaks coinciding with troughs). The least distance change required to cause such a change is a half wavelength distance, so:
λ/2 = 0.03/2
λ = 0.06m
We know,
v = λf
v = 0.06 X 11999Hz
v = 719.94m/s
Therefore, the speed at which sound travels through the gas in the tube is 719.94m/s
Ranking of de Broglie wavelengths from largest to smallest is electron > proton > helium
- De Broglie proposed that because light has both wave and particle properties, matter exhibits both wave and particle properties. This property has been explained as the dual behavior of matter.
- From his observations, de Broglie derived the relationship between the wavelength and momentum of matter. This relationship is known as de Broglie's relationship
De Broglie's relationship is given by
.....(1) , where λ is known as de Broglie wavelength and m is mass , v is velocity , h = Plank’s constant.
From equation (1) wavelength and mass has an inverse relation .
Mass of helium is 4 times the mass of the proton and proton has a greater mass than electron.
According to equation (1) , less the mass higher will be the wavelength
Hence electron having less mass have higher wavelength and then proton and then helium having large mass will have less wavelength .
Thus, order should be electron > proton > helium .
Learn about de brogile wavelength more here :
brainly.com/question/16595523
#SPJ4
Answer:
markers are 29.76 m far apart in the laboratory
Explanation:
Given the data in the question;
speed of particle = 0.624c
lifetime = 159 ns = 1.59 × 10⁻⁷ s
we know that; c is speed of light which is equal to 3 × 10⁸ m/s
we know that
distance = vt
or s = ut
so we substitute
distance = 0.624c × 1.59 × 10⁻⁷ s
distance = 0.624(3 × 10⁸ m/s) × 1.59 × 10⁻⁷ s
distance = 1.872 × 10⁸ m/s × 1.59 × 10⁻⁷ s
distance = 29.76 m
Therefore, markers are 29.76 m far apart in the laboratory
Answer: 15.87 m
From the equation of motion:

where,
is the distance traveled,
is the initial velocity,
is the acceleration and
is the time.
The rock free falls under gravity. Initial velocity,
, 
It took
for rock to hit the water.
Substitute the values in the given equation:

Hence, the water is 15.87 m below the top level of the well.