Answer:
Frequency of sound wave in water = 6.8 Hz
Explanation:
Given:
Speed of sound wave in water = 10.2 m/s
The wavelength of the sound wave in water = 1.5 m
Find:
Frequency of sound wave in water
Computation:
Frequency = Speed of wave / Wavelength
Frequency of sound wave in water = Speed of sound wave in water / The wavelength of the sound wave in water
Frequency of sound wave in water = 10.2 / 1.5
Frequency of sound wave in water = 6.8 Hz
The answer is <span>oligotrophic</span>
Answer:
Answer is explained in the explanation section below.
Explanation:
Solution:
We know from the Coulomb's Law that, Coulomb's force is directly proportional to the product of two charges q1 and q2 and inversely proportional to the square of the radius between them.
So,
F = 
Now, we are asked to get the greatest force. So, in order to do that, product of the charges must be greatest because the force and product of charges are directly proportional.
Let's suppose, q1 = q
So,
if q1 = q
then
q2 = Q-q
Product of Charges = q1 x q2
Now, it is:
Product of Charges = q x (Q-q)
So,
Product of Charges = qQ - 
And the expression qQ -
is clearly a quadratic expression. And clearly its roots are 0 and Q.
So, the highest value of the quadratic equation will be surely at mid-point between the two roots 0 and Q.
So, the midpoint is:
q =
q = Q/2 and it is the highest value of each charge in order to get the greatest force.
Answer:
1.55 m
Explanation:
The potential produced by a point charge, is inversely proportional to the distance from the charge to the point where the potential is being calculated, as follows:

As it only depends from the distance r, we can conclude that if the potential is the same for any point to a distance r from the point charge, the equipotencial surface must be a sphere of radius r.
Replacing q = +1.7*10⁻⁸ C, and k = 9*10⁹ N*m²/C², and V, by 120 V and 54 V, we can find the distance from the charge, to the points where we are calculating the potential V, as follows:


The distance between both points, is just the difference between the radius of both spheres, as follows:
r₂ - r₁ = 1.55 m
When air is blown into the open pipe,
L = 
where nis any integral number 1,2,3,4 etc. and λ is the wavelength of the oscillation
⇒λ=
Note here that n=1 is for fundamental, n=2 is first harmonic and so on..
⇒ third harmonic will be n=4
Given L=6m, n=4, solving for λ we get:
λ=
=3m
Relationship of frequency(f), velocity of sound (c) and wavelength(λ) is:
c=f.λ Or f= 
⇒f=
≈115 Hz