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hjlf
3 years ago
12

Which structure would be most beneficial for a plant in a tundra biome? A. shallow roots that do not reach the frozen soil B. de

ep roots that can access groundwater stores C. thick, tall trunks that release water D. tall, thin trunks that conserve water
Biology
1 answer:
Free_Kalibri [48]3 years ago
6 0

Answer:

The correct answer is<u> A. shallow roots that do not reach the frozen soil</u>

Explanation:

The tundra is known for its lack of trees. The short summer season results in a short growing season, which makes it difficult for trees to grow larger. The cold temperature of the permafrost also makes decomposition slower, which limits the amount of nutrients being cycled through the environment.

Hope this help

plz mark brainliest

Have a nice day!!!

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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

<span>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}
</span>=\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<span> kg^(</span>-1)<span> s^(</span>-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


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