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Katarina [22]
2 years ago
8

A geologist is studying rock layers in an old river bed, and he finds a fossil of a fish and a horsetail rush in the same rock l

ayer. According to the law of faunal and floral succession, the geologist can assume that the rock containing the fossils may date back as far as the _____.
Cambrian period
Ordovician period
Silurian period
Devonian period

Physics
2 answers:
erastovalidia [21]2 years ago
8 0

The correct answer is Devonian period.

This is known to be the period of fishes which occurred 416 million years ago. It is known to be the fourth period of the Paleozoic Era. Significant events like evolution of first insects and other animals.

The Fishes evolved in the Ordovician era but the horses were found in the Devonian period only.

KonstantinChe [14]2 years ago
6 0
A geologist is studying rock layers in an old river bed, and he finds a fossil of a fish and a horsetail rush in the same rock layer. According to the law of faunal and floral succession, the geologist can assume that the rock containing the fossils may date back as far as the <span>Devonian period</span>.
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Explanation:

The parabolic movement results from the composition of a uniform rectilinear motion (horizontal ) and a uniformly accelerated rectilinear motion of upward or downward motion (vertical ).

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Where:  

y: vertical position in meters (m)  

y₀ : initial vertical position in meters (m)  

t : time in seconds (s)

v₀y: initial  vertical velocity  in m/s  

vfy: final  vertical velocity  in m/s  

g: acceleration due to gravity in m/s²

H= hight that reaches the projectile above its starting point (m)

Data

v₀  : total initial speed

θ₀ : angle of v₀ above the horizontal.

g acceleration due to gravity

Calculation of the componentes x-y of the v₀

v₀x = v₀*cos θ₀

v₀y = v₀*sin θ₀

Calculation of the maximum hight that reaches the projectile above its starting point

When the projectile reaches its maximum height (h), vy = 0:

in the Equation (4)

vfy²= v₀y²-2gh

0= v₀y²-2gh

2gh= v₀y²h= \frac{(v_{o})^{2}sin^{2} \theta o }{2g}

2gh=  (v₀*sin θ₀)²

h= \frac{v_{o}^{2} sin^{2} \theta o }{2g}

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