Answer:
The farther star will appear 4 times fainter than the star that is near to the observer.
Explanation:
Since it is given that the luminosity of the 2 stars is same thus they radiate the same energy per unit time
Consider a spherical wave front of energy 'E' that leaves both the stars (Both radiate 'E' as they have same luminosity)
This Energy is spread over the whole surface area of sphere Thus when the wave front is at a distance 'r' the energy per unit surface area is given by
For the star that is twice away from the earth the distance is '2r' thus we will receive an energy given by
Hence we sense it as 4 times fainter than the nearer star.
<span>The velocity would be 54.2 m/s
We would use the equation 1/2mv^2top+mghtop = 1/2mv^2bottom+mghbottom where m is the mass of the bobsled(which can be ignored), vtop/bottom is the velocity of the bobsled at the top or bottom, g is gravity, and htop/bottom is the height of the bobsled at the top or bottom of the hill. Since the velocity of the bobsled at the top of the hill and height at the bottom of the hill are zero, 1/2mv^2top and mghbottom will equal zero. The equation will be mghtop=1/2mv^2bottom. Thus we would solve for v.</span>
It becomes a different element
Answer:
Minimum thickness; t = 9.75 x 10^(-8) m
Explanation:
We are given;
Wavelength of light;λ = 585 nm = 585 x 10^(-9)m
Refractive index of benzene;n = 1.5
Now, let's calculate the wavelength of the film;
Wavelength of film;λ_film = Wavelength of light/Refractive index of benzene
Thus; λ_film = 585 x 10^(-9)/1.5
λ_film = 39 x 10^(-8) m
Now, to find the thickness, we'll use the formula;
2t = ½m(λ_film)
Where;
t is the thickness of the film
m is an integer which we will take as 1
Thus;
2t = ½ x 1 x 39 x 10^(-8)
2t = 19.5 x 10^(-8)
Divide both sides by 2 to give;
t = 9.75 x 10^(-8) m
