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3 years ago

Jawless fish and ocean reefs were devastated by which extinction?

Physics

2 answers:

Iteru [2.4K]3 years ago

7 0

The Late Devonian.

If you go back and reach the text, you may find the passage, "Which species did we lose during this extinction (Late Devonian)? About 20% of all animal families and 70-80% of all animal species were lost. Major victims included the following:

....

- Jawless fish"

Ilia_Sergeevich [38]3 years ago

5 0

Late Devonian (I believe)

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Which of the diagrams below show forces that would result in a movement of the block to the left?

xenn [34]

Diagram A will result in the movement of the block to the left as a result of the forces.

<h3>What is Force?</h3>

This is referred to an influence which is capable of changing the motion of an object.

Diagram A has equal upward and downward force and left side which is 60N is higher than the right side which has 20N. The block will therefore move to the left.

Read more about Force here brainly.com/question/25239010

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2 years ago

A car accelerates from 20.0 m/s to 28.0 m/s over a distance of 50.0 m. What is the car’s acceleration?

Blababa [14]

Answer:

Explanation:

$V^2=V^2_o+2a(x_f-x_i)$

$28^2=20^2+2a(50)\\784=400+100a\\384=100a\\a=3.84m/s^2$

5 0

3 years ago

Please help! :(

Alona [7]

Ep = F * d
F = 20 N
d = 5 m
Ep = 20 N * 5 m
Ep = 100 J

So pick B. A Newton Meter is a joule.

5 0

3 years ago

We know the moon circulates the Earth. Suppose the mass of the Earth and moon are 5.9742 x1024 kg and 7.36 x 1022 kg, whereas th

algol [13]

Answer:

The gravitational force between two objects, one of mass M1 and the other of mass M2, is:

F = G*M1*M2/R^2

Where G is a constant:

G = 6.6x10^11 m^3*/(kg*s^2)

And R is the distance between the two objects.

M1 = 5.9742x10^24 kg

M2 = 7.36x10^22 kg

R = 382171 km = 382171000 m

Then the gravitational force is:

F = 6.6x10^-11 m^3*/(kg*s^2)*(5.9742x10^24 kg)*(7.36x10^22 kg)/(382171000 m)^2

F = 1.987x10^20 N

4 0

4 years ago

A boy exerts a force of 9.0 N horizontally to push his sister on a sled. He pushes her through a distance of 15 m. How much work

MrRa [10]

Answer:

Work done by the boy is 135 J and is positive work.

Friction is the force acting that stops the sled and her sister.

Work done by friction is negative work.

Explanation:

Given:

Force acting on the sled by the boy is, $F=9.0\ N$

Displacement of the sled in the direction of force is, $S=15\ m$

We know that, positive work is said to be done by a force, if the force causes displacement in its own direction. Also, negative work is said to be done by a force, if the force causes displacement in the direction exactly opposite to the direction of the applied force.

Here, the force applied by the boy causes a displacement in the direction of the applied force. So, work is positive and is given as:

Work = Force × Displacement

$Work=F\times S\\Work=9\times 15=135\ J$

Therefore, the work done by the boy is positive and equal to 135 J.

Now, when the force of push is removed, then the body will experience only the frictional force in the opposite direction which eventually causes the sled to stop.

Here, the direction of force is backward while the displacement is in the forward direction. So, both of them are in opposite direction.

So, work done by frictional force is negative.

6 0

4 years ago