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3 years ago

An athlete runs four laps of a 400 m track. What is the athlete's total displacement?

Physics

1 answer:

sasho [114]3 years ago

7 0

Answer:

○ 0 $\checkmark$

Explanation:

Even if the athelete runs four laps of a 400 m track, its displacement will be 0 because the displacement is the shortest distance from its to final position. And, here the final and initial position is same since he comes to its initial position after covering certain distance. So, displacement is 0.

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Particle A and particle B, each of mass M, move along the x-axis exerting a force on each other. The potential energy of the sys

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Speed of particle B is 2v₀/3 m/s to the left. Particle A and particle B will always have equal speed since they experience equal forces.

<h3>Conservation of energy</h3>

The speed and direction of the particle B is determined by applying the principle of conservation of energy as follows;

K.E₁ + P.E₁ = K.E₂ + P.E₂

$\frac{1}{2} Mv^2_A + \frac{G}{r^2} = \frac{1}{2} Mv^2_B + \frac{G}{r^2} \\\\ \frac{1}{2} Mv^2_A = \frac{1}{2} Mv^2_B\\\\v^2_A = v^2_B\\\\v_A = v_B$

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The complete question is below:

Particle A and particle B, each of mass M, move along the x-axis exerting a force on each other. The potential energy of the system of two particles assosicated with the force is given by the equation U=G/r 2, where r is the distance between the two particles and G is a positive constant. At time t=T1 particle A is observed to be traveling with speed 2vo/3 to the left. The speed and direction of motion of particle B is ?

Learn more about conservation of energy here: brainly.com/question/166559

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Answer:

The speed of the package of mass m right before the collision $= 7.668\ ms^-1$

Their common speed after the collision $= 2.56\ ms^-1$

Height achieved by the package of mass m when it rebounds $= 0.33\ m$

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

where $K$ is Kinetic energy and $U$ is Potential energy.

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Considering the fact $K_{initial} = 0\ and U_{final} =0$ we will plug out he values of the given terms.

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Keypoints:

Sum of energies and momentum are conserved in all collisions.
Sum of KE and PE is also known as Mechanical energy.
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<u>KE = Kinetic Energy and PE = Potential Energy</u>

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

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Cancelling m from both sides and dividing 3 on both sides.

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c.Now the collision is perfectly elastic $e=1$

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$V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1$

Now using law of conservation of energy.

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An athlete runs four laps of a 400 m track. What is the athlete's total displacement?​

An athlete runs four laps of a 400 m track. What is the athlete's total displacement?