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Debora [2.8K]
3 years ago
11

Which of the following is the best paraphrasing of the Heisenberg uncertainty principle?

Physics
1 answer:
nadezda [96]3 years ago
7 0
<h2>Answer: The more precisely you know the position of a particle, the less well you can know the momentum of the particle </h2>

The Heisenberg uncertainty principle was enunciated in 1927. It postulates that the fact that each particle has a wave associated with it, imposes restrictions on the ability to determine <u>its position and speed at the same time.  </u>

In other words:  

<em>It is impossible to measure simultaneously (according to quantum physics), and with absolute precision, the value of the position and the momentum (linear momentum) of a particle.</em>  

<h2>So, the greater certainty is seeked in determining the position of a particle, the less is known its linear momentum and, therefore, its mass and velocity.  </h2><h2 />

In fact, even with the most precise devices, the uncertainty in the measurement continues to exist. Thus, in general, the greater the precision in the measurement of one of these magnitudes, the greater the uncertainty in the measure of the other complementary variable.  

Therefore the correct option is C.

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Answer:

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Zina [86]

Answer:

The speed of the 8-ball is 2.125 m/s after the collision.

Explanation:

<u>Law Of Conservation Of Linear Momentum</u>

The total momentum of a system of masses is conserved unless an external force is applied. The momentum of a body with mass m and velocity v is calculated as follows:

P=mv

If we have a system of masses, then the total momentum is the sum of all the individual momentums:

P=m_1v_1+m_2v_2+...+m_nv_n

When a collision occurs, the velocities change to v' and the final momentum is:

P'=m_1v'_1+m_2v'_2+...+m_nv'_n

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m_1v_1+m_2v_2=m_1v'_1+m_2v'_2

The m1=0.16 Kg 8-ball is initially at rest v1=0. It is hit by an m2=0.17 Kg cue ball that was moving at v2=2 m/s.

After the collision, the cue ball comes to rest v2'=0. It's required to find the final speed v1' after the collision.

The above equation is solved for v1':

\displaystyle v'_1=\frac{m_1v_1+m_2v_2-m_2v'_2}{m_1}

\displaystyle v'_1=\frac{0.16*0+0.17*2-0.17*0}{0.16}

\displaystyle v'_1=\frac{0.34}{0.16}

v'_1=2.125\ m/s

The speed of the 8-ball is 2.125 m/s after the collision.

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