Answer:
Added sugars, Saturated fat, Trans fat.
Explanation:
Power = Work done/Time taken
So, keeping this in mind,we can solve it as follows:
= 700/3.1
= 7000/31
= 225.80 W
Answer:
<em>The correct answer is (a) 312.5nm and (b) 125nm</em>
Explanation:
<em>The first step to take is to find The minimum thickness of the slick of the oil</em>
<em>Given that,</em>
<em>(a) tmin λ/2n </em>
<em>We substitute 750nm ( in air) for λ and 1.20 for n for the expression of minimum thickness t of the oil slick at that spot</em>
<em>thus,</em>
<em>tmin = (750nm)/2(1.2) = 312.5nm</em>
<em>The minimum thickness of the oil slick at that spot is =312.5nm</em>
<em>(B) we find the minimum thickness t </em>
<em>The minimum thickness of the oil slick at the spot will be calculated by,</em>
<em>tmin = λ/4n</em>
<em>we then 750nm ( in air) for λ and 1.50 for n in the expression for the minimum thickness of the slick of the oil.</em>
<em>tmin = (750nm)/4 (1.5) = 125nm</em>
<em>Therefore the minimum thickness t will now be = 125nm</em>
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Answer:
Explanation:
Let T be the tension in the string and mg is the weight of the block.
T = mg
By using Newton's second law
T - mg = ma
mg - mg = ma
a = 0
Thus, the acceleration is zero.
Answer:
Explanation:
The original has hybrid 15N/14N DNA, and the second generation has both hybrid 15N/14N DNA and 14N/14N DNA. No 15N/15N DNA was observed. In this experiment:
Nitrogen is a significant component of DNA. 14N is the most bounteous isotope of nitrogen, however, DNA with the heavier yet non-radioactive and 15N isotope is likewise practical.
E. coli was developed for several generations in a medium containing NH4Cl with 15N. When DNA is extracted from these cells and centrifuged on a salt density gradient, the DNA separates at which its density equals to the salt arrangement. The DNA of the cells developed in 15N medium had a higher density than cells developed in typical 14N medium. After that, E. coli cells with just 15N in their DNA were transferred to a 14N medium.
DNA was removed and compared to pure 14N DNA and 15N DNA. Immediately after only one replication, the DNA was found to have an intermediate density. Since conservative replication would result in equal measures of DNA of the higher and lower densities yet no DNA of an intermediate density, conservative replication was eliminated. Moreso, this result was consistent with both semi-conservative and dispersive replication. Semi conservative replication would result in double-stranded DNA with one strand of 15N DNA, and one of 14N DNA, while dispersive replication would result in double-stranded DNA with the two strands having mixtures of 15N and 14N DNA, either of which would have appeared as DNA of an intermediate density.
The DNA from cells after two replications had been completed and found to comprise of equal measures of DNA with two different densities, one corresponding to the intermediate density of DNA of cells developed for just a single division in 14N medium, the other corresponding to DNA from cells developed completely in 14N medium. This was inconsistent with dispersive replication, which would have resulted in a single density, lower than the intermediate density of the one-generation cells, yet at the same time higher than cells become distinctly in 14N DNA medium, as the first 15N DNA would have been part evenly among all DNA strands. The result was steady with the semi-conservative replication hypothesis. The semi conservative hypothesis calculates that each molecule after replication will contain one old and one new strand. The dispersive model suggests that each strand of each new molecule will possess a mixture of old and new DNA.
