Answer:
You take the light from a star, planet or galaxy and pass it through a spectroscope, which is a bit like a prism letting you split the light into its component colours. "It lets you see the chemicals being absorbed or emitted by the light source. From this you can work out all sorts of things," says Watson
Answer:
Y = V / f where Y equals wavelength
4 Y1 = V / f1 for a closed pipe the wavelength is 1/4 the length of the pipe
2 Y2 = V / f2 for the open pipe the wavelength is 1/2 the length of the pipe
Y1 / Y2 = 2 = f2 / f1 dividing equations
f2 = 2 f1
the new fundamental frequency is 2 * 130.8 = 261.6
(The new wavelength is 1/2 the original wavelength so the frequency must double to produce the same speed.
Answer:
388.97 nm
Explanation:
The computation of the wavelength of this light in benzene is shown below:
As we know that
n (water) = 1.333
n (benzene) = 1.501
And, the wavelength of water is 438 nm
![\lambda (benzene) = \lambda (water) [\frac{n(water)}{n(benzene}]](https://tex.z-dn.net/?f=%5Clambda%20%28benzene%29%20%3D%20%5Clambda%20%28water%29%20%5B%5Cfrac%7Bn%28water%29%7D%7Bn%28benzene%7D%5D)
Now placing these values to the above formula
So,
= 388.97 nm
We simply applied the above formula so that we can easily determine the wavelength of this light in benzene could come
Answer:
I = M R^2 is the moment of inertia about a point that is a distance R from the center of mass (uniform distributed mass).
The moment of inertia about the center of a sphere is 2 / 5 M R^2.
By the parallel axis theorem the moment of inertia about a point on the rim of the sphere is I = 2/5 M R^2 + M R^2 = 7/5 M R^2
I = 7/5 * 20 kg * .2^2 m = 1.12 kg m^2
To develop this problem it is necessary to apply the concepts related to a magnetic field in spheres.
By definition we know that the magnetic field in a sphere can be described as
Where,
a = Radius
z = Distance to the magnetic field
I = Current
Permeability constant in free space
Our values are given as
diameter of the sphere then,
Thus z = a
Re-arrange to find I,
Therefore the current at the pole of this sphere is 
