Answer:
the stronger light 5.5 m apart from the total illumination
Explanation:
From the problem's statement , the following equation can be deducted:
I= k/r²
where I = intensity of illumination , r= distance between the point and the light source , k = constant of proportionality
denoting 1 as the stronger light and 2 as the weaker light
I₁= k/r₁²
I₂= k/r₂²
dividing both equations
I₂/I₁ = r₁²/r₂²=(r₁/r₂)²
solving for r₁
r₁ = r₂ * √(I₂/I₁)
since we are on the line between the two light sources , the distance from the light source to the weaker light is he distance from the light source to the stronger light + distance between the lights . Thus
r₂ = r₁ + d
then
r₁ = (r₁ + d)* √(I₂/I₁)
r₁ = r₁*√(I₂/I₁) + d*√(I₂/I₁)
r₁*(1-√(I₂/I₁)) = d*√(I₂/I₁)
r₁ = d*√(I₂/I₁)/(1-√(I₂/I₁)) =
r₁ = d/[√(I₁/I₂)-1)]
since the stronger light is 9 times more intense than the weaker
I₁= 9*I₂ → I₁/I₂ = 9 →√(I₁/I₂)= 3
then since d=11 m
r₁ = d/[√(I₁/I₂)-1)] = 11 m / (3-1) = 5.5 m
r₁ = 5.5 m
therefore the stronger light 5.5 m apart from the total illumination
Answer:
If each element can be identified by its spectrum then the composition of an unknown star can be determined
Explanation:
The chemical nature of the elements is that they absorb specific wavelength of light depending on their atom. By spectral analysis of the spectrum of emitted light by a body, the body's composition can therefore be determined. As such in order to determine the composition of distant bodies such as planets, stars and other celestial bodies scientists usually make use of spectroscopy.
The average atomic mass of Jz is 331.3 u.
The average atomic mass of Jz is the <em>weighted average</em> of the atomic masses of its isotopes.
We multiply the atomic mass of each isotope by a number representing its <em>relative importance</em> (i.e., its fractional abundance).
Thus,
⅗ × 329.1 u = 197.46 u
⅖ × 334.5 u = <u>133.80 u</u>
TOTAL = 331.3 u
Answer:
atom, ions, protons neutrons, atomic number
Explanation:
Answer: single replacement
Explanation: metals higher than hydrogen in the activity series will always displaced or replace hydrogen in a solution. Pb is higher than hydrogen in the activity series, so it will displaces or replace hydrogen in a solution