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Gwar [14]
2 years ago
7

Elastic and Inelastic Collisions​

Physics
1 answer:
Igoryamba2 years ago
4 0

\large{\color{gold}{\bigstar }}\:  \: { \underline{\mathfrak{Together \: We \: Go \: Far}}}  \:  \:   \color{gold}\bigstar

\rule{70mm}{2.9pt}

<h3>Elastic and Inelastic Collisions</h3>

A collision is an interaction between two objects resulting in the exchange of impulse and momentum. The time of impact is usually small; the impulse provided by external forces like friction during this time is negligible. When two or more bodies collide, the momentum of the system is therefore approximately conserved. In an isolated system, the total momentum before the collision is equal to the total momentum of the system after collision.

total momentum before collision = total momentum after collision

There are two types of collision and are categorized according to whether the system's total kinetic energy changes. It may or may not be conserved depending on the type of collision. It may lose during collisions when (1) it is converted to heat or other forms like binding energy, sound, light (if there is spark), etc. and (2) it is spent in producing deformation or damage, such as when two cars collide.

-  -  -  -  -  -  -  -  -  -  -  - -   \: -

The two types of collisions are:-

1. Elastic collision ☞ states that the system's total kinetic energy does not change and colliding objects bounce off after collision. (no kinetic energy is loss, no damage, no heat)

Examples:

  • Motion of atoms and molecules
  • Hitting billiard balls
  • When a soccer player kicks a ball since the player's foot and the ball do indeed remain completely separate after collision.

2. Inelastic collision ☞ states that the system's total kinetic energy changes (i.e., converted to some other form of energy). Collision is said to be perfectly inelastic if after collision the objects stick together and move as one mass with one velocity.

Example:

  • Celestial bodies collide, like two asteroids, they fuse together to form a larger body.

=  =  =  =  =  =  =  =  =  =  =  =  =  =  = = = =

Additional Information:-

☞ brainly.com/question/24616147

☞ brainly.com/question/12941951

\rule{70mm}{2.9pt}

~ \large {\color{iron}{(:}} \:  \bold {MrHarold  \: \color{iron}{ :)}}

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I WILL GIVE BRAINLIEST IF SOMEONE GETS THIS......
pav-90 [236]

Answer:

Explanation:

a)

Firstly to calculate the total mass of the can before the metal was lowered we need to add the mass of the eureka can and the mass of the water in the can. We don't know the mass of the water but we can easily find if we know the volume of the can. In order to calculate the volume we would have to multiply the area of the cross section by the height. So we do the following.

100cm^{2} x 10cm = 1000cm^{3}

Now in order to find the mass that water has in this case we have to multiply the water's density by the volume, and so we get....

\frac{1g}{cm^{3} } x 1000cm^{3} = 1000g or 1kg

Knowing this, we now can calculate the total mass of the can before the metal was lowered, by adding the mass of the water to the mass of the can. So we get....

1000g + 100g = 1100g or 1.1kg

b)

The volume of the water that over flowed will be equal to the volume of the metal piece (since when we add the metal piece, the metal piece will force out the same volume of water as itself, to understand this more deeply you can read the about "Archimedes principle"). Knowing this we just have to calculate the volume of the metal piece an that will be the answer. So this time in order to find volume we will have to divide the total mass of the metal piece by its density. So we get....

20g ÷ \frac{8g}{cm^{3} } = 2.5 cm^{3}

c)

Now to find out the total mass of the can after the metal piece was lowered we would have to add the mass of the can itself, mass of the water inside the can, and the mass of the metal piece. We know the mass of the can, and the metal piece but we don't know the mass of the water because when we lowered the metal piece some of the water overflowed, and as a result the mass of the water changed. So now we just have to find the mass of the water in the can keeping in mind the fact that 2.5cm^{3} overflowed. So now we the same process as in number a) just with a few adjustments.

\frac{1g}{cm^{3} } x (1000cm^{3} - 2.5cm^{3}) = 997.5g

So now that we know the mass of the water in the can after we added the metal piece we can add all the three masses together (the mass of the can. the mass of the water, and the mass of the metal piece) and get the answer.

100g + 997.5g + 20g = 1117.5g or 1.1175kg

5 0
3 years ago
2. In the pendulum lab, the period (swing time) was O a. the independent variable O b.graphed on the vertical-axis O c.graphed o
Nadya [2.5K]
B is appropriate answer
7 0
3 years ago
I need the answer right now pls help me
yaroslaw [1]

The AREA of the shaded region is the moving object's displacement.

7 0
4 years ago
This term means that we are interacting with an item at a high-level, with lower-level internal details hidden from the user.
katovenus [111]

Option d) abstraction is the correct answer.

Abstraction is the process of taking away or removing characteristics from something in order to reduce it to a set of essential characteristics. In object-oriented programming, abstraction is one of three central principles Through this process, a programmer hides all but the relevant data about an object in order to reduce complexity and increase efficiency.

Learn more about Abstraction here:

brainly.com/question/19419456

#SPJ4

7 0
2 years ago
A muon is a short-lived particle. It is found experimentally that muonsmoving at speed 0.9ccan travel about 1500 meters between
Monica [59]

Answer:

a)t=5.5\mu s

b)\Delta t'=12.5\mu s

Explanation:

a) Let's use the constant velocity equation:

v=\frac{\Delta x}{\Delta t}

  • v is the speed of the muon. 0.9*c
  • c is the speed of light 3*10⁸ m/s

t=\frac{\Delta x}{v}=\frac{1500}{0.9*(3*10^{8})}

t=5.5\mu s

b) Here we need to use Lorentz factor because the speed of the muon is relativistic. Hence the time in the rest frame is the product of the Lorentz factor times the time in the inertial frame.

\Delta t'=\gamma\Delta t

\gamma =\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

v is the speed of muon (0.9c)

Therefore the time in the rest frame will be:

\Delta t'=\frac{1}{\sqrt{1-\frac{(0.9c)^{2}}{c^{2}}}}\Delta t

\Delta t'=\frac{1}{\sqrt{1-0.9^{2}}}\Delta t

\Delta t'=\frac{1}{0.44}\Delta t

No we use the value of Δt calculated in a)

\Delta t'=2.27*5.5=12.5\mu s

I hope it helps you!

     

 

8 0
4 years ago
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