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erastovalidia [21]
3 years ago
9

The two particles are both moving to the right. Particle 1 catches up with particle 2 and collides with it. The particles stick

together and continue on with velocity vf. Which of these statements is true?
A. vf=v1
B. vf=v2
C. vf is less than v2
D. vf is less than v2 but greater than v1
E. vf>v1

Physics
1 answer:
Irina18 [472]3 years ago
8 0

Answer:

See bolded below.

Explanation:

Consider the " Before " and " After. " " Before, " this particle 1 was trying to catch up with this particle 2, and " after " particle one had collided with particle two. Take a look at the attachment below for a more detailed examination.

Here is how this will play out. Particle 1, with great velocity, will hit particle 2, which would mean that Particle 2 has less velocity than Particle 1. Now after the collision, energy is transferred to Particle 2, and while Particle 1 has now stopped in it's tracks, Particle 2 - with more energy than before - will continue as long as it has to before friction eventually brings it to a stop.

_______________________________________________________

From this we can conclude that Vf, from the picture below, must have less energy than V1, but more energy than V2 - and vice versa.

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vovikov84 [41]

This question involves the concept of kinetic energy.

The student's claim is "right".

<h3>Kinetic Energy</h3>

The energy possessed by a body, by the virtue of its motion is called kinetic energy. Mathematically it is given by the following formula:

K.E =\frac{1}{2}mv^2

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Therefore,

For the paintball:

K.E = \frac{1}{2}(4\ g)(90\ m/s)^2

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K.E = \frac{1}{2}(1\ g)(180\ m/s)^2

K.E = 16200 J

Hence, both paintball and pellet will have same kinetic energy. The student is right.

Learn more about kinetic energy here:

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2 years ago
Models are frequently used in science to assist in the understanding of complex information. Models can include items like a wor
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Three point charges are located on the x-axis. The first charge, q1 = 10 μC, is at x = -1.0 m. The second charge, q2 = 20 μC, is
victus00 [196]

Answer:

<em>3.15 N towards the positive x-axis</em>

<em></em>

Explanation:

first charge has charge q1 = 10 μC = 10 x 10^-6 C

second charge has charge q2 = 20 μC = 20 x 10^-6 C

third charge has charge q3 = -30 μC = -30 x 20^-6 C

According to coulomb's law, force between two charged particle is given as

F = \frac{-kQq}{r^2}

Where

F is the force between the charges

k is Coulomb's constant = 9 x 10^9 kg⋅m^3⋅s^−2⋅C^−2.

Q is the magnitude of one charge

q is the magnitude of the other charge

is the distance between these two charges

For the force on q2 due to q1,

distance r between them = 0 - (-1.0) = 1 m

F = \frac{-9*10^{9}*10*10^{-6}*20*10^{-6}}{1^2} = -1.8 N (the negative sign indicates a repulsion on q2 towards the positive  x-axis)

For the force on q2 due to q3,

distance between them = 2.0 - 0 = 2 m

F = \frac{-9*10^{9}*20*10^{-6}*(-30*10^{-6})}{2^2} = 1.35 N (the positive sign indicates an attraction on q2 towards the positive x-axis)

Resultant force on q2 = 1.8 N + 1.35 N = <em>3.15 N towards the positive x-axis</em>

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3 years ago
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artcher [175]
We have all the charges for q1, q2, and q3. 
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F(1) = F (2on1) + F (3on1)

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F(2on1) = 3.37 N

Since F1 is 7N,

F(1) = F (2on1) + F (3on1)
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Since it wil be going in the negative direction,
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F(3on1) = -10.37N

F(3on1) = k |q1 q3| / r(the distance between the two)^2 
r^2 x F(3on1) = k |q1 q3| 
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So it's located in -.144m

Thank you for posting your question. I hope that this answer helped you. Let me know if you need more help. 
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Answer:

496.7 K

Explanation:

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