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3 years ago

What provides bones with flexibility and tensile strength

Biology

2 answers:

lilavasa [31]3 years ago

8 0

calcium is the answer

erastovalidia [21]3 years ago

5 0

Collagen fibrils arranged in lamellae results in the tensile strength of the bone.

With love,

Hope :)

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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

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

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What provides bones with flexibility and tensile strength ​

What provides bones with flexibility and tensile strength