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2 years ago

What types of animals are prosimians?

Biology

1 answer:

valentina_108 [34]2 years ago

8 0

Prosimians are a group of less-developed primates.

<h3>What are Prosimians?</h3>

They are a group of primates with some less-developed features as compared to other primates.

The more developed primates are referred to as Simians and these include humans, apes, and monkeys.

Prosimians include both living and extinct animals such as the lemur, lorises, and galagos.

More on Prosimians can be found here: brainly.com/question/14521463

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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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Answer:

Explanation:

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The correct answer is -

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