Identify the names of animals that lived on Earth before dinosaurs.

Physics

2 answers:

Mumz [18]2 years ago

7 0

Im sure there are others but i know that bugs existed as long as the earth has

Minchanka [31]2 years ago

7 0

Mainly sea creatures such as scorpions, jellyfish, and some other pre-historic creatures. Hope this helps.

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Can you list the offensive position on a flag football team?

Alik [6]

Answer:

yes u can flag football has everything that pad football has so you can enlist on being offensive position but you have to play like you want that position

Explanation:

6 0

2 years ago

What is the mass of a stone moving at a speed of 15 m/s and having a monument at 7.1 kg meters per second

adell [148]

Answer:

<h3>The answer is 0.47 kg</h3>

Explanation:

The mass of the object given it's momentum and velocity can be found by using the formula

$m = \frac{p}{v} \\$

where

p is the momentum

v is the velocity

We have

$m = \frac{7.1}{15} \\ = 0.4733333...$

We have the final answer as

Hope this helps you

4 0

2 years ago

a 42.3 kg girl and a 7.93 kg sled are on the surface of a frozen lake, 15.0m apart and linked by a rope, but not moving yet. the

ycow [4]

Answer:

they meet from the girl's original position at: 2.37 (meters)

Explanation:

We need to use the Newton's law, exactly the second law that relate force, mass and acceleration as: $F=m*a$ with this we can get both accelerations; solving for acceleration $a=\frac{F}{m}$ . Now $a_{girl}=\frac{5.76}{42.3}=0.14 (m/s^{2})$ and $a_{sled}=\frac{5.76}{7.93}=0.73(m/s^{2})$ . Then knowing that they both travel at the same time and assuming that the distance among the girl and the sled is: 15.0-x, so, $x=\frac{1}{2}*a_{girl}*t^{2}$ and $15.0-x=\frac{1}{2}*a_{sled}*t^{2}$ , solving for the time we get: $t=\sqrt{\frac{2x}{a_{girl} } }$ and $t=\sqrt{\frac{2*(15.0-x)}{a_{sled} } }$ with this equations we solving for the x that is the distance between the girl and the sled after the apply the force and we get: $\sqrt{\frac{2x}{a_{girl}}} = \sqrt{\frac{2*(15.0-x)}{a_{sled} }$ . Finally we get: $\frac{x}{a_{girl} }=\frac{(15.0-x)}{a_{sled} }$ and replacing the values we have got: $\frac{x}{0.14} =\frac{(15.0-x)}{0.73}$ so $5.33*x=15-x$ so x=2.37 (meters).