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Fynjy0 [20]
3 years ago
9

In this diagram of a long bone, which type of bone marking is circled?

Biology
2 answers:
Juliette [100K]3 years ago
7 0

The answer is c osteon

JulsSmile [24]3 years ago
4 0
The answer is Osteon.  Osteons<span> are several millimeters long and about 0.2 millimeter in diameter; they tend to run parallel to the long axis of a bone. The </span>osteon<span>units of bone are made up of Haversian canals (HC) and Volkmann canals (VC), which run perpendicular to the long axes of </span>osteons<span> and connect adjacent Haversian canals.</span>
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As the ball is dropped from rest V_{1O} = 0, so:

(2) V_{1f} = - gt

Note that the velocity is going to be negative as the ball is going down. To get the time it would take the ball to reach de base you can use the kinematic equation for position:

(3) h_{1} = h_{1O} + V_{1O} *t - \frac{1}{2} gt^{2}

We need the answer when h=0, and from the initial conditions V_{1O} = 0, h_{1O} = 24m, you get:

(4) 24m = \frac{1}{2} gt_{1f}^{2}

Solving for t, with 9,8\frac{m}{s^{2}} and :

(5) t_{1f} = \sqrt{\frac{24m*2}{g} } = 2,21 s

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(6) V_{1f} = -21,66  \frac{m}{s}

So the initial velocity of ball 2 is equial to this but oppossite in direction so: V_{2O}= -V_{1f} = 21,66  \frac{m}{s}

The general position equation for ball 2 is (considering h_{2O} = 0 :

(7) h_{2} = V_{2O} *t - \frac{1}{2} gt^{2}

They cross paths when h_{1}=h_{2} so:

(8) h_{1O} - \frac{1}{2} gt^{2} = V_{2O} *t - \frac{1}{2} gt^{2}

Rearranging:

(9) t_{cross} =\frac{h_{1O}}{V_{2O} }

Replacing values:

t_{cross} = 1,108 s

To get the absolute position replace t_{cross} on equation (3) or (7):

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To get it below the top of te cliff:

h_{cross, from above} = 24 m - 17,98 m

h_{cross, from above} = 6,02 m

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