Answer:
The remaining percentage of drug concentration is about 88.7% 2 years after manufacture.
Explanation:
Recall the formula for the decay of a substance at an initial concentration at manufacture:
where k is the decay rate (in our case 0.06/year), and t is the elapsed time in years. Therefore, after 2 years since manufacture we have:
This in percent form is 88.7 %. That is, the remaining percentage of drug concentration is about 88.7% 2 years after manufacture.
Answer:
1.696 nm
Explanation:
For a diffraction grating, dsinθ = mλ where d = number of lines per metre of grating = 5510 lines per cm = 551000 lines per metre and λ = wavelength of light = 467 nm = 467 × 10⁻⁹ m. For a principal maximum, m = 1. So,
dsinθ = mλ = (1)λ = λ
dsinθ = λ
sinθ = λ/d.
Also tanθ = w/D where w = distance of center of screen to principal maximum and D = distance of grating to screen = 1.03 m
From trig ratios 1 + cot²θ = cosec²θ
1 + (1/tan²θ) = 1/(sin²θ)
substituting the values of sinθ and tanθ we have
1 + (D/w)² = (d/λ)²
(D/w)² = (d/λ)² - 1
(w/D)² = 1/[(d/λ)² - 1]
(w/D) = 1/√[(d/λ)² - 1]
w = D/√[(d/λ)² - 1] = 1.03 m/√[(551000/467 × 10⁻⁹ )² - 1] = 1.03 m/√[(1179.87 × 10⁹ )² - 1] = 1.03 m/1179.87 × 10⁹ = 0.000848 × 10⁻⁹ = 0.848 × 10⁻¹² m = 0.848 nm.
w is also the distance from the center to the other principal maximum on the other side.
So for both principal maxima to be on the screen, its minimum width must be 2w = 2 × 0.848 nm = 1.696 nm
So, the minimum width of the screen must be 1.696 nm
Answer:
A metalloid is used because it is a semiconductor and can become more conductive when more light shines on it
Explanation:
The material used in a solar panel is a metalloid. It can often become conductive when more light shines on it.
Metalloids have properties that straddles between those of metals and non-metals.
In essence, they can be conductive or not under certain conditions.
The most important property they exhibit is that they can become more conductive when more light shines on them. This way more electrons are produced.
Answer:
The importance of the sediments permeability is that if it is permeable, water will flow easily through the sediment and thereby produce a very good supply of water for the well.
Explanation:
When digging a well into saturated sediments, the possibility of the sediment with either little saturation or full saturation being able to provide steady water supply will be limited by how permeable it is. Now, the importance of the sediments permeability is that if it is permeable, water will flow easily through the sediment and thereby produce a very good supply of water for the well.
Yes that looks correct to me. good luck!!