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3 years ago

Are dams density-dependent or density-independent limiting factors?

Biology

2 answers:

alexdok [17]3 years ago

8 0

Answer:

Density-independent

Explanation:

konstantin123 [22]3 years ago

7 0

The answer is density-independent. Hope this helps!

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Both flat worms in round worms have a body type called?

olga2289 [7]

should be called Bilateral symmetry

6 0

3 years ago

A hypothetical phylogeny for marsupial relatedness is shown here. Macropodidae is the marsupial family. Which of these statement

Darina [25.2K]

Answer: B and E

Explanation: USATESTPREP

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2 years ago

How do some nucleotide sequences in DNA get copied into RNA when DNA and RNA have different bases. For example, how is the seque

kvasek [131]

Answer:

In RNA the base thymine is essencially replaced by the base uracil. So the sequence TGAC in DNA would get copied into RNA as ACUG.

Adenine pairs with uracil in RNA like it pairs with thymine in DNA

Explanation:

8 0

2 years ago

What often happens to rocks that undergo chemical weathering?

Schach [20]

Breaks down over time and becomes dirt

4 0

3 years ago

Mars had an orbital period of 1.88 years. In two or more complete sentences,explain how to calculate the average distance from m

Vanyuwa [196]

This is an interesting problem involving astronomy, in fact, simple physics.
Let r=distance of sun to mars, in metres

Mars had an orbital period of 1.88 years.
=>
tangential velocity, v, of the planet, in m/s is
$v=\frac{2\pi{r}}{T}$
 $=\frac{2\pi{r}}{1.88*365.25*86400}$ m/s, accounting for leap years
$=3.371*10^{-8}\pi{r}$ m/s

The centripetal force, Fc, generated is
$Fc=\frac{mv^2}{r}$
where m=mass of mars = 6.39*10^(24) kg
$=\frac{mv^2}{r}$
$=\frac{6.39*10^{24}v^2}{r}$
$=7.26168*10^9\pi^2r$

The gravitation pull from the sun, Fg, is given by
$Fg=\frac{GMm}{r^2}$
where G=grav. const., =6.67408*10^(-11) m^3 kg^(-1) s^(-2)
M=mass of sun=1.989*10^(30) kg
$=\frac{6.67408*10^{-11}1.989*10^{30}6.39*10^{24}}{r^2}$
$=\frac{8.4826*10^44}{r^2}$

Since the radial distance is in equilibrium, the average distance, r can be found by equation Fc=Fg and solving for r:
Fc=Fg
=>
$7.26168*10^9\pi^2r=\frac{8.4826*10^44}{r^2}$
Solving for the real root:
$r^3=\frac{8.48256*10^44}{7.26168*10^9*%pi^2}$
$=\frac{1.1681263*10^{35}}{\pi^2}$
=2.279*10^11 m

5 0

3 years ago