All three windows are the same size.
A has 10 complete waves visible through the window. B has 3, and C has 4.
So A must have the smallest wavelengths.
A glass pipe system has a very corrosive liquid flowing in it (think hydrofluoric acid, say). The liquid will destroy flow meters, but you need to know the flow rate. One way of measuring the flow rate is to add a fluorescent dye to the liquid at a known concentration, and then downstream activate the dye by UV light and then measure the dye concentration by emitted light. If the dye is added at 1.00 g/s, and the dye concentration downstream is 0.050% by mass, what is the unknown flow rate in kg/h
glass
The answer is a.12.5kg because i just did the test and it was correct.
hope this helps
Answer:
Hipparchus was an ancient Greek who classified stars based on the brightness in 129 B.C. He grouped the brightest stars and ranked them 1 (first magnitude) and dimmest stars as 6 (sixth magnitude). Thus, the smaller numbers indicated brighter stars. Now, the scale extends in negative axis as well. More the negative number, brighter is the star. For example, Sun has magnitude -26.74.
This the apparent magnitude which means the classification is based on the brightness of the star as it appears from the Earth.
A) We want to find the work function of the potassium. Apply this equation:
E = 1243/λ - Φ
E = energy of photoelectron, λ = incoming light wavelength, Φ = potassium work function
Given values:
E = 2.93eV, λ = 240nm
Plug in and solve for Φ:
2.93 = 1243/240 - Φ
Φ = 2.25eV
B) We want to find the threshold wavelength, i.e. find the wavelength such that the energy E of the photoelectrons is 0eV. Plug in E = 0eV and Φ = 2.25eV and solve for the threshold wavelength λ:
E = 1243/λ - Φ
0 = 1243/λ - Φ
0 = 1243/λ - 2.25
λ = 552nm
C) We want to find the frequency associated with the threshold wavelength. Apply this equation:
c = fλ
c = speed of light in a vacuum, f = frequency, λ = wavelength
Given values:
c = 3×10⁸m/s, λ = 5.52×10⁻⁷m
Plug in and solve for f:
3×10⁸ = f(5.52×10⁻⁷)
f = 5.43×10¹⁴Hz
