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olasank [31]
3 years ago
7

How is the axe an example of a wedge (think about simple machines)?

Physics
1 answer:
vfiekz [6]3 years ago
5 0

Answer:

Heavier on one side lighter on the other side.

:P

Explanation:

You might be interested in
A small ball of mass 2.00 kilograms is moving at a velocity 1.50 meters/second. It hits a larger, stationary ball of mass 5.00 k
rewona [7]

The kinetic energy of the small ball before the collision is

                             KE  =  (1/2) (mass) (speed)²

                                     = (1/2) (2 kg) (1.5 m/s)

                                     =    (1 kg)  (2.25 m²/s²)

                                     =        2.25 joules.

Now is a good time to review the Law of Conservation of Energy:

                     Energy is never created or destroyed. 
                     If it seems that some energy disappeared,
                     it actually had to go somewhere.
                     And if it seems like some energy magically appeared,
                     it actually had to come from somewhere.

The small ball has 2.25 joules of kinetic energy before the collision.
If the small ball doesn't have a jet engine on it or a hamster inside,
and does not stop briefly to eat spinach, then there won't be any
more kinetic energy than that after the collision.  The large ball
and the small ball will just have to share the same 2.25 joules.

3 0
4 years ago
Using the graph above, explain what happened on the amount carbon in the atmosphere 1960 to 2020.
Ivenika [448]

Answer:

No graph...

Explanation:

4 0
3 years ago
Mathew throws a ball straight up in the air. it rises for a period of time and then begins to drop. At which points in the balls
Allushta [10]

Answer:   The greatest force of gravity on the ball will occur at the point when the ball is near to hit the ground ... hope this helps have a good day :P

3 0
3 years ago
What is the electric potential v due to the nucleus of hydrogen at a distance of 5.292×10−11 m ?
viktelen [127]

Answer: 27.21 V

Explanation:

The <u>electric potential</u> V_{E} due to a point charge is expressed as:

V_{E}=k\frac{q}{r}

Where:

k=9(10)^{9}\frac{Nm^{2}}{C^{2}} is the <u>electric constant</u>

q=1.6(10)^{-19}C is the <u>electric charge of the hydrogen nucleus</u>, which is positive

r=5.292(10)^{-11}m is the <u>distance</u>

Rewritting the equation with the known values:

V_{E}=9(10)^{9}\frac{Nm^{2}}{C^{2}}\frac{1.6(10)^{-19}C}{5.292(10)^{-11}m}

Finally:

V_{E}=27.21 V

5 0
3 years ago
Brainliest if correct Question 7 of 10
OleMash [197]

Answer:

I would think A

Explanation:

4 0
3 years ago
Read 2 more answers
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