Explanation:
Suppose you want to shine a flashlight beam down a long, straight hallway. Just point the beam straight down the hallway -- light travels in straight lines, so it is no problem. What if the hallway has a bend in it? You could place a mirror at the bend to reflect the light beam around the corner. What if the hallway is very winding with multiple bends? You might line the walls with mirrors and angle the beam so that it bounces from side-to-side all along the hallway. This is exactly what happens in an optical fiber.
The light in a fiber-optic cable travels through the core (hallway) by constantly bouncing from the cladding (mirror-lined walls), a principle called total internal reflection. Because the cladding does not absorb any light from the core, the light wave can travel great distances.
However, some of the light signal degrades within the fiber, mostly due to impurities in the glass. The extent that the signal degrades depends on the purity of the glass and the wavelength of the transmitted light (for example, 850 nm = 60 to 75 percent/km; 1,300 nm = 50 to 60 percent/km; 1,550 nm is greater than 50 percent/km). Some premium optical fibers show much less signal degradation -- less than 10 percent/km at 1,550 nm.
1
Answer:
Explanation:
Given that there are two force of 1 pound each at right angles to each other.
The from the vector law of addition:
where:
resultant force
be the two of the forces to be added.
The pH of the buffer solution is 7.30 for 0.172 m in Hypochlorous acid and 0.131 m in Sodium hypochlorite.
<h3>What is a buffer solution?</h3>
A weak acid and its conjugate base, or a weak base and its conjugate acid, are mixed together to form a solution called a buffer solution, which is based on water as the solvent. They do not change in pH when diluted or when modest amounts of acid or alkali are added to them. A relatively little amount of a strong acid or strong base has little effect on the pH of buffer solutions. As a result, they are employed to maintain a steady pH.
According to the question:
Ka = 3.8×10⁻⁸
pKa = - log (Ka)
= - log(3.8×10⁻⁸)
= 7.42
pH = pKa + log {[conjugate base]/[acid]}
= 7.42+ log {0.131/0.172}
= 7.302
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Answer:
the major difference between the rate of diffusion between urea and albumin is that urea diffused more slowly because it is larger than albumin.
Explanation:
As described in activity 1, the major difference between the rate of diffusion between urea and albumin is that urea diffused more slowly because it is larger than albumin. Urea basically is an heavy compound, thus it is at the base of the chain reactions which which major purpose is to break down the amino acids that make up proteins. Albumin on the other hand is a form of protein that simply helps in the transmission of blood and prevent blood from flowing through to other tissue in the body. The liver produces the albumin the body. So going back to the question, the major difference between the rate of diffusion between urea and albumin is that urea diffused more slowly because it is larger than albumin.
Answer:
ΔP = 986 Kpa
Explanation:
The solution is given in the pictures below
