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alex41 [277]
3 years ago
10

Two balls, each with a mass of 0.5 kg, collide on a pool table. Is the law of conservation of momentum satisfied in the collisio

n? Explain why or why not
Physics
2 answers:
Dmitriy789 [7]3 years ago
7 0

During the collision between two balls on the pool table there is no external force along the line of collision between them

Since there is no external force on it so here we can say

F = 0 = \frac{\Delta P}{\Delta t}

here we have

\Delta P = 0

so we can say

P_i = P_f

since there is no external force so we can say during the collision the momentum of two balls will remain conserved

aleksklad [387]3 years ago
7 0
<h3>Sample Response: <em>Yes, the law of conservation of momentum is satisfied. The total momentum before the collision is 1.5 kg • m/s and the total momentum after the collision is 1.5 kg • m/s. The momentum before and after the collision is the same.</em></h3><h3><em /></h3>

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When waves strike an object and bounce off, what has occured
Vesna [10]
The answer is reflection
5 0
3 years ago
Find the angle of prism if the ray just fail to emerge from 2nd face When ray of light falls normallly on the face of prism of R
Usimov [2.4K]

The angle of prism is 41.81 degrees.

<u>Explanation:</u>

For no emergence to be taken place, inside a prism, Total Internal Reflection (TIR) should take place at the second surface.  For TIR, at second surface, angle of refraction must be greater than critical angle. Angle of prism is related to refraction as,

                            A>r_{1}+C

Since, r_{1} = C and A \geq 2 C

This implies A \geq C

                         \sin A \geq \sin C

                         \sin A \geq \frac{1}{\mu}

                         \sin A \geq \frac{2}{3}

when sin goes to other side become as sin inverse of value, and obtain the result as below,

                       A=41.81^{\circ}

3 0
3 years ago
A soccer ball is kicked from the top of one building with a height of H1 = 30.2 m to another building with a height of H2 = 12.0
viktelen [127]

Hi there!

Initially, we have gravitational potential energy and kinetic energy. If we set the zero-line at H2 (12.0m), then the ball at the second building only has kinetic energy.

We also know there was work done on the ball by air resistance that decreased the ball's total energy.

Let's do a summation using the equations:
KE = \frac{1}{2}mv^2 \\\\PE = mgh

Our initial energy consists of both kinetic and potential energy (relative to the final height of the ball)

E_i = \frac{1}{2}mv_i^2 + mg(H_1 - H_2)

Our final energy, since we set the zero-line to be at H2, is just kinetic energy.

E_f = \frac{1}{2}mv_f^2

And:
W_A = E_i - E_f

The work done by air resistance is equal to the difference between the initial energy and the final energy of the soccer ball.

Therefore:
W_A = \frac{1}{2}mv_i^2 + mg(H_1 - H_2) -  \frac{1}{2}mv_f^2

Solving for the work done by air resistance:
W_A = \frac{1}{2}(.450)(15.1^2)+ (.450)(9.8)(30.2 - 12) -  \frac{1}{2}(.450)(19.89^2)

W_A = \boxed{42.552 J}

8 0
2 years ago
Road rage is an aggressive driving incident where the driver has:
WARRIOR [948]
I would say that it would be A.) Been insulted. This is because you are angered by someone else. It does not necessarily mean that something has de-escelated, you've lost control of your car, or you've nothing to lose. Hopefully that helps. :)

5 0
3 years ago
A ship's wheel has a moment of inertia of 0.930 kilogram·meters squared. The inner radius of the ring is 26 centimeters, and the
Vikki [24]

We can use the formula of the moment of inertia given by:

r\cdot F=I\alpha

Where:

r = Distance from the point about which the torque is being measured to the point where the force is applied

F = Force

I = Moment of inertia

α = Angular acceleration

So:

\begin{gathered} r\cdot F=(-0.26\times314+290\times0.32)=92.8-81.64=11.16 \\ I=0.930 \\ so,_{\text{ }}solve_{\text{ }}for_{\text{ }}\alpha: \\ \alpha=\frac{r\cdot F}{I} \\ \alpha=\frac{11.16}{0.930} \\ \alpha=\frac{12rad}{s^2} \end{gathered}

Answer:

12 rad/s²

8 0
1 year ago
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