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3 years ago

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A particle undergoes a constant acceleration of 3.90 m/s2. After a certain amount of time, its velocity is 12.2 m/s. (Where appl

icable, indicate the direction with the sign of your answer.) (a) If its initial velocity is 6.1 m/s, what is its displacement during this time?

1 answer:

Law Incorporation [45]3 years ago

8 0

Answer:

$s_{f} = 14.312 m$

Explanation:

Since the particle is experimenting a constant acceleration, the displacement can be found by using this formula:

$v_{f}^{2} = v_{o}^{2} + 2 \cdot a \cdot (s_{f} - s_{o})$

Since $s_{o} = 0 m$ , the equation is simplified to this form:

$v_{f}^{2} = v_{o}^{2} + 2 \cdot a \cdot s_{f}$

Then, the displacement is now isolated:

$s_{f} = \frac{v_{f}^{2}-v_{o}^{2}}{2 \cdot a}$

Terms are replaced herein:

$s_{f} = \frac{(12.2 \frac{m}{s})^{2}-(6.1 \frac{m}{s})^{2}}{2 \cdot (3.90 \frac{m}{s^{2}}) } \\s_{f} = 14.312 m$

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The electric force exerted on a charged particle is given by

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For a positive charge, the direction of the force is the same as the electric field.

In this problem:

$q=+4.9\mu C=+4.9\cdot 10^{-6}C$ is the charge

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So the electric force (along the x-direction) is:

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The magnetic force instead is given by

$F=qvB sin \theta$

where

q is the charge

v is the velocity of the charge

B is the magnetic field

$\theta$ is the angle between the directions of v and B

Here the charge is stationary: this means $v=0$ , therefore the magnetic force due to each component of the magnetic field is zero.

b)

In this case, the particle is moving along the +x axis.

The magnitude of the electric force does not depend on the speed: therefore, the electric force on the particle here is the same as in part a,

$F_{E_x}=1.19\cdot 10^{-3} N$ (towards positive x-direction)

Concerning the magnetic force, we have to analyze the two different fields:

- $B_x$ : this field is parallel to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=0^{\circ}$ , so the force due to this field is zero.

$- B_y$ : this field is perpendicular to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=90^{\circ}$ . Therefore, $\theta=90^{\circ}$ , so the force due to this field is:

$F_{B_y}=qvB_y$

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Substituting,

$F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N$

And the direction of this force can be found using the right-hand rule:

- Index finger: direction of the velocity (+x axis)

- Middle finger: direction of the magnetic field (+y axis)

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$F_{B_y}=qvB_y$

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And by using the right-hand rule:

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