Answer:
The new frequency (F₂ ) will be related to the old frequency by a factor of one (1)
Explanation:
Fundamental frequency = wave velocity/2L
where;
L is the length of the stretched rubber
Wave velocity = 
Frequency (F₁) = 
To obtain the new frequency with respect to the old frequency, we consider the conditions stated in the question.
Given:
L₂ =2L₁ = 2L
T₂ = 2T₁ = 2T
(M/L)₂ = 0.5(M/L)₁ = 0.5(M/L)
F₂ = ![\frac{\sqrt{\frac{2T}{0.5(\frac{M}{L})}}}{4*L} = \frac{\sqrt{4(\frac{T}{\frac{M}{L}}})}{4*L} = \frac{2}{2} [\frac{\sqrt{\frac{T}{\frac{M}{L}}}}{2*L}] = F_1](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B%5Cfrac%7B2T%7D%7B0.5%28%5Cfrac%7BM%7D%7BL%7D%29%7D%7D%7D%7B4%2AL%7D%20%3D%20%5Cfrac%7B%5Csqrt%7B4%28%5Cfrac%7BT%7D%7B%5Cfrac%7BM%7D%7BL%7D%7D%7D%29%7D%7B4%2AL%7D%20%3D%20%5Cfrac%7B2%7D%7B2%7D%20%5B%5Cfrac%7B%5Csqrt%7B%5Cfrac%7BT%7D%7B%5Cfrac%7BM%7D%7BL%7D%7D%7D%7D%7B2%2AL%7D%5D%20%3D%20F_1)
Therefore, the new frequency (F₂ ) will be related to the old frequency by a factor of one (1).
The answer would be B, George Darwin :)
Answer:
E=0
Explanation:
The electric field at the centre of the shell is zero because total enclosed charge in the nucleus is zero
A) 1.55
The speed of light in a medium is given by:
where
is the speed of light in a vacuum
n is the refractive index of the material
In this problem, the speed of light in quartz is
So we can re-arrange the previous formula to find n, the index of refraction of quartz:
B) 550.3 nm
The relationship between the wavelength of the light in air and in quartz is
where
is the wavelenght in quartz
is the wavelength in air
n is the refractive index
For the light in this problem, we have
Therefore, we can re-arrange the equation to find , the wavelength in air:
Answer:
Explanation:
The two charges are q and Q - q. Let the distance between them is r
Use the formula for coulomb's law for the force between the two charges
So, the force between the charges q and Q - q is given by
For maxima and minima, differentiate the force with respect to q.
For maxima and minima, the value of dF/dq = 0
So, we get
q = Q /2
Now 
the double derivate is negative, so the force is maxima when q = Q / 2 .
