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tresset_1 [31]
3 years ago
8

Can someone explain the law of conservation of energy?

Physics
1 answer:
Bad White [126]3 years ago
5 0
The law of conservation of energy is:

-- Energy can't be created or destroyed.

-- Energy can't just appear out of nowhere.  If you suddenly have
more energy, then the 'extra' energy had to come from somewhere.

-- Energy can't just disappear.  If you suddenly have less energy,
then the 'missing' energy had to go somewhere.
________________________________________

There are also conservation laws for mass and electric charge.
They say exactly the same thing.  Just write 'mass' or 'charge'
in the sentences up above, in place of the word 'energy'.
________________________________________

And now I can tell you that the conservation laws for energy and mass
are actually one single law ... the conservation of mass/energy.  That's
because we discovered about 100 years ago that mass can convert
into energy, and energy can convert into mass, and it's the total of BOTH
of them that gets conserved (can't be created or destroyed).

How much mass makes how much energy ?

The answer is        E = m c²  .

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If a ball of mass, M, moving at velocity, v, collided with a ball of mass 10M at rest, describe what could happen to the velocit
KatRina [158]
Refer to the diagram shown below.

Before collision, the momentum of the two masses is
P₁ = Mv + (10M)*0 = Mv

After the collision, assume that the lighter ball rebounds off the heavier ball with a coefficient of restitution of r, so that v₂ = rv.
If r = 1, the rebound is elastic and v₂ = -v.
If r < 1, the rebound velocity is v₂ = -rv.
If r= 0, the lighter ball sticks to the heavier ball.

The momentum after collision is
P₂ = -Mv₂ + 10Mv₁

Because momentum is conserved, P₁ = P₂. That is,
10Mv₁ - M(rv) = Mv
v₁ = v(1+r)/10  for r>0.

When r=1 (elastic rebound)
v₁ = v/5.
The heavier ball moves right at 20% of the velocity of the lighter ball,
and the lighter ball rebounds with its velocity in the opposite direction.

When 0 < r < 1,
v₁ = (1+r)/10.
The heavier ball travels with greater than 20% of the velocity of the lighter ball, and the lighter ball rebounds with a velocity less than its initial velocity.

When r=0, the balls will stick together and
(10M + M)v₁ = Mv
v₁ = v/11.
The stuck balls move together at 1/11 of the initial velocity of the lighter ball.


3 0
3 years ago
What are some potential sources of error in the<br> experiment that is described in the article?
mezya [45]

Explanation:

common source of error include instrumental, environmental, procedural , and human Police Stop all of these errors can be either random for systematic depending on how they affect the results.

4 0
3 years ago
when a beam of light passes at an oblique angle into the material of lower optical density the angle is of incidence is
butalik [34]

Penn Foster Students: less than the angle of refraction

3 0
3 years ago
If it takes an amount of work W to move two q point charges from infinity to a distance d apart from each other, then how much w
aleksklad [387]

Answer:3W

If it takes an amount of work W to move two q point charges from infinity to a distance d apart from each other, then how much work should it take to move three q point charges from infinity to a distance d apart from each other?

A) 2W

B) 3W

C) 4W

D) 6W

Explanation: calculating work done,W, in moving two positive q point charges from infinity to a valued distance d from each other  is

W = k(+q)(+q)/ d

k is couloumb's constant

work done in moving 3 equal positive charges from infinity to a finite distance is given by

W₂=W₄=W₆=k(+q)(+q)/ d

Total work done, W' =k(+q)(+q)/ d + k(+q)(+q)/ d + k(+q)(+q)/ d

= W + W + W = 3W

7 0
3 years ago
If the object represented by the FBD below has a mass of 2.5 kg, what is the acceleration of the object?
Debora [2.8K]

Answer:

4 m/s² down

Explanation:

We'll begin by calculating the net force acting on the object.

The net force acting on the object from the left and right side is zero because the same force is applied on both sides.

Next, we shall determine the net force acting on the object from the up and down side. This can be obtained as follow:

Force up (Fᵤ) = 15 N

Force down (Fₔ) = 25 N

Net force (Fₙ) =?

Fₙ = Fₔ – Fᵤ

Fₙ = 25 – 15

Fₙ = 10 N down

Finally, we shall determine the acceleration of the object. This can be obtained as follow:

Mass (ml= 2.5 Kg

Net force (Fₙ) = 10 N down

Acceleration (a) =?

Fₙ = ma

10 = 2.5 × a

Divide both side by 2.5

a = 10 / 2.5

a = 4 m/s² down

Therefore, the acceleration of the object is 4 m/s² down

6 0
3 years ago
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