Explanation: Roughly 180 - 200 million years ago, just before the first dinosaurs evolved. Mammals themselves evolved from a group or reptiles which exhibited mammal-like traits. One of them was specialized teeth. Reptiles tend to have teeth all the same shape. The mammal-like reptiles evolved tiny teeth in front of the jaw and two pairs of over sized fangs along the the sides. Like modern mammals, the head was large in proportion to the rest of the body. The jaws were also evolving another mammal trait, the ability to move sideways. Despite the lack of specialized teeth, acute hearing and the ability to chew, the dinosaurs evolved an adaptation which made them far more successful than mammals--modified leg bones. These limbs could be articulated directly under their bodies. This enabled the legs to support more weight, since the limbs were now under the body instead of at the sides. Then dinosaurs did something which secured their dominance for the next 120 million years - they began to stand on two legs. Although the back was still parallel to the ground, running on two legs greatly increased the dinosaur's speed. Mammals could simply not compete with swift, giant predators and were forced to remain small, and most became nocturnal to evade dinosaurs which were probably active during the day. Despite that they managed to survive which allowed the further evolution of mammals into us, humans.

5 0

3 years ago

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Describe the composition of the nucleus of an atom.

gavmur [86]

Answer:

Explanation:the atom consists of a tiny nucleus at its center which is surrounded by a moving electrons. The nucleus contains a positively charged proton equal in size with the negatively charged electrons . The nucleus also may contain neutrons which have the same mass with the protons but no charge is neutral.

4 0

3 years ago

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A 100-kg running back runs at 5 m/s into a stationary linebacker. It takes 0.5 s for the running back to be completely stopped.

Elza [17]

Answer:

1000 N

Explanation:

First, we need to find the deceleration of the running back, which is given by:

$a=\frac{v-u}{t}$

where

v = 0 is his final velocity

u = 5 m/s is his initial velocity

t = 0.5 s is the time taken

Substituting, we have

$a=\frac{0-5 m/s}{0.5 s}=-10 m/s^2$

And now we can calculate the force exerted on the running back, by using Newton's second law:

$F=ma=(100 kg)(-10 m/s^2)=-1000 N$

so, the magnitude of the force is 1000 N.

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3 years ago

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A hollow cylinder that is rolling without slipping is given a velocity of 5.0 m/s and rolls up an incline to a vertical height o

inysia [295]

Answer:

The hollow cylinder rolled up the inclined plane by 1.91 m

Explanation:

From the principle of conservation of mechanical energy, total kinetic energy = total potential energy

$M.E_T = \frac{1}{2}mv^2 + \frac{1}{2} I \omega^2 + mgh$

The total energy at the bottom of the inclined plane = total energy at the top of the inclined plane.

$\frac{1}{2}mv_i^2 + \frac{1}{2} I \omega_i^2 + mg(0) = \frac{1}{2}mv_f^2 + \frac{1}{2} I \omega_f^2 + mgh$

moment of inertia, I, of a hollow cylinder = ¹/₂mr²

substitute for I in the equation above;

$\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}mr^2 \omega_i^2) = \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}mr^2 \omega_f^2) + mgh\\\\ but \ v = r \omega\\\\\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}m v_i^2 ) = \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}m v_f^2) + mgh\\\\\frac{1}{2}mv_i^2 +\frac{1}{4}mv_i^2 = \frac{1}{2}mv_f^2 +\frac{1}{4}mv_f^2 +mgh\\\\\frac{3}{4}mv_i^2 = \frac{3}{4}mv_f^2 +mgh\\\\mgh = \frac{3}{4}mv_i^2 - \frac{3}{4}mv_f^2\\\\gh = \frac{3}{4}v_i^2 - \frac{3}{4}v_f^2\\\\$

$h = \frac{3}{4g}(v_1^2 -v_f^2)$

given;

v₁ = 5.0 m/s

vf = 0

g = 9.8 m/s²

$h = \frac{3}{4g}(v_1^2 -v_f^2) =\frac{3}{4*9.8}(5^2 -0) = 1.91 \ m$

Therefore, the hollow cylinder rolled up the inclined plane by 1.91 m

5 0

3 years ago