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ziro4ka [17]
3 years ago
12

Molecules like glucose and amino acids are not lipid soluble, so cannot cross the cell membrane by simple diffusion. Instead, th

ey are shuttled across by carrier proteins in the membrane from high to low concentration. Does this process require energy from the cell? This is an example of what type of molecular movement?
Biology
1 answer:
schepotkina [342]3 years ago
3 0

Answer: this is called passive transport and it does not require energy.

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A family doctor may prescribe this anti-invective medicine to treat pneumonia.
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What is the weakest bone in your body?​
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<u><em>stapes</em></u>

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Explanation:

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What is one reason why many geologists did not at first accept the theory of continental drift?
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The first reason was that his theory of continental drift was just too weak for most geologists to accept. Even though he believed the supercontinent that broke up into different continents moved, he did not have a clear explanation to how the continents moved. The other reason is that some of his explanation clashed with ideas that were widely accepted in the science communities. He used similar fossils from different continents to back up his theory of continental drift. However, at that time, many scientists that had observed similarities in fossils in places like South America and Africa believed there were similar fossils in different continents because of a land bridge that were formed by two continents.


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Which statement correctly describes the rock cycle?
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g If a person takes a prescribed dose of 10 milligrams of Valium, the amount of Valium in that person's bloodstream at any time
Elenna [48]

Answer:

a. 8.1 milligrams

b. 40.07 hours

c. 8.859 milligrams

Explanation:

If a person takes a prescribed dose of 10 milligrams of Valium, the amount of Valium in that person's bloodstream at any time can be modeled by

A_{t}=10e^{-0.0173t}

Where A(t) = amount of Valium remaining in the blood after t hours

t = time or duration after the drug is taken

a. we have to calculate the amount of drug remaining in the bloodstream after 12 hours

A_{12}=10e^{-0.0173\times12}

A_{12}=10e^{-0.2076}

                = 10×0.81253

                = 8.1 milligrams

b. In this part we have to calculate the time when A(t) = 5 milligrams

5=10e^{-0.0173\timest}

\frac{5}{10}=e^{-0.0173t}

0.5 = e^{-0.0173t}

Now we take natural log on both the sides of the equation.

ln(0.5) = ln(e^{-0.0173t})

-0.69314 = -0.0173t

t = \frac{0.69314}{0.0173}

t = 40.0658

 ≈ 40.07 hours

c. In this part we have to calculate the rate, by which amount of drug will decay in the bloodstream after 7 hours.

A_{7}=10e^{-0.0173\times7}

A_{7}=10e^{-0.1211}

              = 10×0.8859

              = 8.859 milligrams

6 0
4 years ago
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