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3 years ago

5

This cell would be in a (blank) solution

1 answer:

Natali5045456 [20]3 years ago

7 0

Answer:

hey you didn't provide the complete question. what's the full question maybe I could help

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Elden [556K]

Answer:

A

Explanation:

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4 years ago

How are the number of chromosomes maintained during asexual reproduction?

Harrizon [31]

Answer:

<em>The number of chromosomes is maintained through the process of mitosis during asexual reproduction. </em>

Mitosis can be described as a process in which a cell divides and produces two daughter cells which are exactly similar to the parent cell. The chromosome number of the daughter cells will be the same as the parent cells. No crossing over or recombination occurs during this type of cell division. Hence, mitosis is an effective way to maintain the chromosome number of organisms which reproduce asexually.

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3 years ago

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 The process in which the DNA from one type of cell causes an inheritable change in a second type of cell is called

frosja888 [35]

Answer:

bacteriophages

Explanation:

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3 years ago

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A fossil is discovered that has only 12.5% of the carbon-14 that it would have had originally (when the animal was alive). If th

mariarad [96]

Answer:

The fossil is 17,100 years old.

Explanation:

The decay equation:

$\frac{dN}{dt}\propto -N$

$\Rightarrow \frac{dN}{dt}= -\lambda N$

$\Rightarrow \frac{dN}{N}= -\lambda dt$

Integrating both sides

$\Rightarrow \int\frac{dN}{N}= \int-\lambda dt$

$\Rightarrow ln |N|=-\lambda t+c$

When t=0, N= $N_0$ = initial amount

$ln |N_0|=-\lambda .0+c$

$\Rightarrow ln |N_0|=c$

$\therefore ln |N|=-\lambda t+ln|N_0|$

$\Rightarrow ln |N|-ln|N_0|=-\lambda t$

$\Rightarrow ln |\frac {N}{N_0}|=-\lambda t$

$\Rightarrow \frac {N}{N_0}=e^{-\lambda t}$

$\Rightarrow N=N_0e^{-\lambda t}$

The decay equation is

$N=N_0e^{-\lambda t}$

Given that,

The half life of carbon - 14 is 5700 years.

For half life, $N=\frac{1}{2} N_0$

To find the value of $\lambda$ , we need to put the value of N and t in the decay equation.

$\frac12N_0=N_0e^{-\lambda \times 5700}$

$\Rightarrow \frac12=e^{-\lambda \times 5700}$ [ Divided $N_0$ both sides]

Taking ln both sides

$\Rightarrow ln| \frac12|=ln|e^{-\lambda \times 5700}|$

$\Rightarrow ln| \frac12|={-\lambda \times 5700}$

$\Rightarrow \lambda= \frac{ln| \frac12|}{-5700}$

$\Rightarrow \lambda= \frac{ln|1|-ln|2|}{-5700}$ [ $ln|\frac mn|= ln |m|-ln |n|$ ]

$\Rightarrow \lambda= \frac{ln|2|}{5700}$ [ln 1= 0]

The fossil has only 12.5% of the carbon carbon-14 that it would have had originally.

So, $N=\frac{12.5}{100} N_0$

Then,

$\frac{12.5}{100} N_0=N_0e^{-\frac{ln|2|}{5700}t$

$\Rightarrow \frac{12.5}{100} =e^{-\frac{ln|2|}{5700}t$

Taking ln both sides

$\Rightarrow ln|\frac{12.5}{100} |=ln|e^{-\frac{ln|2|}{5700}t}|$

$\Rightarrow ln|\frac{12.5}{100} |={-\frac{ln|2|}{5700}t}$

$\Rightarrow t=\frac{ ln|\frac{12.5}{100}|} {-\frac{ln|2|}{5700}}$

$\Rightarrow t=\frac{ ln|\frac{12.5}{100}|\times 5700} {-{ln|2|}}$

$\Rightarrow t=17,100$

The fossil is 17,100 years old.

7 0

4 years ago

Why do you think so many objects are able to impact the moons surface

tigry1 [53]

Answer: there is no gravity and no type of life on the moon therefore basically making it a huge target.

Explanation:

8 0

3 years ago