Which situations describe an elastic collision?

Physics

2 answers:

Charra [1.4K]4 years ago

4 0

Answer;

A. two glass marbles bounce off each other.

Explanation;

Elastic collisions are collisions in which both momentum and kinetic energy are conserved. The total system kinetic energy before the collision equals the total system kinetic energy after the collision. If total kinetic energy is not conserved, then the collision is referred to as an inelastic collision.

An elastic collision occurs when the two objects "bounce" apart when they collide. Two rubber balls are a good example. In an elastic collision, both momentum and kinetic energy are conserved. Almost no energy is lost to sound, heat, or deformation.

Elan Coil [88]4 years ago

3 0

Elastic collisions are collisions in which both momentum and kinetic energy are conserved. I think the correct answer from the choices listed above is option A. Two glass marbles bounce off each other is an example of an elastic collision. Hope this answers the question.

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If you ran 15 km/h for 20 min, how much distance would you cover?

nikitadnepr [17]

Since we have 15 kilometers per hour, and we're looking for 20 minutes, let's set up proportions.
20/60 minutes = x/15
20/60 = 1/3, so let's leave that simplified.
1/3 = x/15
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1 • 5 = 5, so you will cover 5 kilometers in 20 minutes.

I hope this helps!

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Two identical small metal spheres with q1 > 0 and |q1| > |q2| attract each other with a force of magnitude 81 mN when sepa

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Answer:

Explanation:

Check the attachment for solution

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3 years ago

Suppose you first walk 12.0 m in a direction 20? west of north and then 20.0 m in a direction 40.0? south of west. how far are y

Gnesinka [82]

The representation of this problem is shown in Figure 1. So our goal is to find the vector $\overrightarrow{R}$ . From the figure we know that:

$\left | \overrightarrow{A} \right |=12m \\ \\ \left | \overrightarrow{B} \right |=20m \\ \\ \theta_{A}=20^{\circ} \\ \\ \theta_{B}=40^{\circ}$

From geometry, we know that:

$\overrightarrow{R}=\overrightarrow{A}+\overrightarrow{B}$

Then using vector decomposition into components:

$For \ A: \\ \\ A_x=-\left | \overrightarrow{A} \right |sin\theta_A=-12sin(20^{\circ})=-4.10 \\ \\ A_y=\left | \overrightarrow{A} \right |cos\theta_A=12cos(20^{\circ})=11.27 \\ \\ \\ For \ B: \\ \\ B_x=-\left | \overrightarrow{B} \right |cos\theta_B=-20cos(40^{\circ})=-15.32 \\ \\ B_y=-\left | \overrightarrow{B} \right |sin\theta_B=-20sin(40^{\circ})=-12.85$

Therefore:

$R_x=A_x+B_x=-4.10-15.32=-19.42m \\ \\ R_y=A_y+B_y=11.27-12.85=-1.58m$

So if you want to find out how far are you from your starting point you need to know the magnitude of the vector $\overrightarrow{R}$ , that is:

$\left | \overrightarrow{R} \right |=
\sqrt{R_x^2+R_y^2}=\sqrt{(-19.42)^2+(-1.58)^2}=\boxed{19.48m}$

Finally, let's find the compass direction of a line connecting your starting point to your final position. What we are looking for here is an angle that is shown in Figure 2 which is an angle defined with respect to the positive x-axis. Therefore:

 $\theta_R=180^{\circ}+tan^{-1}(\frac{\left | R_y \right |}{\left | R_x \right |}) \\ \\ \theta_R=180^{\circ}+tan^{-1}(\frac{1.58}{19.42}) \\ \\ \theta_R=180^{\circ}+4.65^{\circ}=185.85^{\circ}$

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3 years ago

6. Draw a velocity-time graph for an object originally traveling at -3 m/s. The object

faltersainse [42]

See the graph in attachment

Explanation:

In this problem we have to draw a velocity-time graph for an object travelling initially at -3 m/s, then slowing down and turning around.

In the graph, we see that the initial velocity at time t = 0 is

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and it is negative, so below the x-axis.

Later, the object slows down: this means that the magnitude of its velocity increases, therefore (since the velocity is negative) the curve must go upward, approaching and reaching the x-axis (which corresponds to zero velocity).

After that, the object's velocity keep increasing, but now it is positive: this means that the object is travelling in a direction opposite to the initial direction, so it has turned around.

Learn more about velocity:

brainly.com/question/5248528

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