When there are 14c-lable uracil that are added to the growth medium of cells, the macromolecules that will be labled are RNA. Uracil is a nucleobase that make up the DNA or the RNA. In RNA, uracil binds with other nucleobase (adenine) through hydrogen bonds.
Answer:
passively diffuses down its concentration gradient through the endothelial cell plasma membrane out of the cell and then passiveley diffuses through the plasma membrane into the cytoplasm of the smooth muscle cell, where it acts to decrease contraction.
Explanation:
Hello,
At first, we must consider that 
and 
, passively diffuses through membranes. As it is produced by an enzyme and accumulates in the endothelial cell cytosol, passively diffuses down its concentration gradient through the endothelial cell plasma membrane out of the cell and then passiveley diffuses through the plasma membrane into the cytoplasm of the smooth muscle cell, where it acts to decrease contraction.
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Answer:
76,6 kg
Explanation:
A kg it's equal to 1x10^3 grams
A Gigagrams it's equal to 1x10^9 grams
Knowing this, a kg it's equal to 1x10^6 gigagrams
![7,66*10^{-5}[gigagram]*\frac{1*10^6 [kg]}{1 [gigagram]}= 76.6 [kg]](https://tex.z-dn.net/?f=7%2C66%2A10%5E%7B-5%7D%5Bgigagram%5D%2A%5Cfrac%7B1%2A10%5E6%20%5Bkg%5D%7D%7B1%20%5Bgigagram%5D%7D%3D%2076.6%20%5Bkg%5D)
Answer:
16 minutes
Explanation:
First, we need to calculate the amount of heat needed to cool the beef stew:
Q = mcΔT
Where <em>m</em> is the mass, <em>c</em> is the heat capacity and <em>ΔT</em> is the variation of the temperature.
Q = 10x4x(40 - 90)
Q = -2000 kJ
So, the beef stew needs to lost 2000 kJ to cool.
With the initial temperature at 90ºC, the rate of cooling(r) will be:
r = 1.955x(90 - 25)
r = 127.075 kJ/min
So, to lose 2000 kJ, will be necessary:
t = Q/r
t = 2000/127.075
t = 16 minutes
