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4 years ago

9

A particle executes simple harmonic motion with an amplitude of 3.00 cm. At what position does its speed equal half of is maximu

m speed?

1 answer:

Mrac [35]4 years ago

5 0

Answer:

x=±0.026m

Explanation:

In simple harmonic motion the maximum value of the magnitude of velocity

$v_{max}=wA=\sqrt{\frac{k}{m} }A$

The speed as a function of position for simple harmonic oscillator is given by

$v=w\sqrt{A^{2}-x^{2}$

where A is amplitude of motion

Given data

Amplitude A=3 cm =0.03 m

v=(1/2)Vmax

To find

We have asked to find position x does its speed equal half of is maximum speed

Solution

The speed of the particle the maximum speed as:

$v=\frac{V_{max} }{2}\\ w\sqrt{A^{2}-x^{2} }=\frac{wA}{2}\\ A^{2}-x^{2}=\frac{A^{2} }{4}\\ x^{2}=A^{2}- \frac{A^{2} }{4}\\ x^{2}=\frac{3A^{2}}{4} \\x=\sqrt{\frac{3A^{2}}{4}}$

x=±(√3(0.03)/2)

x=±0.026m

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A transparent oil with index of refraction 1.28 spills on the surface of water (index of refraction 1.33), producing a maximum o

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3 years ago

Describe what happens in a crash according to newton's first law

Nostrana [21]

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The law of inertia is most commonly experienced when riding in cars and trucks. In fact, the tendency of moving objects to continue in motion is a common cause of a variety of transportation injuries - of both small and large magnitudes. Consider for instance the unfortunate collision of a car with a wall. Upon contact with the wall, an unbalanced force acts upon the car to abruptly decelerate it to rest. Any passengers in the car will also be decelerated to rest if they are strapped to the car by seat belts. Being strapped tightly to the car, the passengers share the same state of motion as the car. As the car accelerates, the passengers accelerate with it; as the car decelerates, the passengers decelerate with it; and as the car maintains a constant speed, the passengers maintain a constant speed as well.

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3 years ago

zubka84 [21]

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That's all I can answer for you. Hope it helps

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4 years ago

Charge q1 = +2.00 μC is at -0.500 m along the x axis. Charge q2 = -2.00 μC is at 0.500 m along the x axis. Charge q3 = 2.00 μC i

Kobotan [32]

The magnitude of <em>electrical</em> force on charge $q_{3}$ due to the others is 0.102 newtons.

<h3>How to calculate the electrical force experimented on a particle</h3>

The vector <em>position</em> of each particle respect to origin are described below:

$\vec r_{1} = (-0.500, 0)\,[m]$

$\vec r_{2} = (+0.500, 0)\,[m]$

$\vec r_{3} = (0, +0.500)\,[m]$

Then, distances of the former two particles particles respect to the latter one are found now:

$\vec r_{13} = (+0.500, +0.500)\,[m]$

$r_{13} = \sqrt{\vec r_{13}\,\bullet\,\vec r_{13}} = \sqrt{(0.500\,m)^{2}+(0.500\,m)^{2}}$

$r_{13} =\frac{\sqrt{2}}{2}\,m$

$\vec r_{23} = (-0.500, +0.500)\,[m]$

$r_{23} = \sqrt{\vec r_{23}\,\bullet \,\vec r_{23}} = \sqrt{(-0.500\,m)^{2}+(0.500\,m)^{2}}$

$r_{23} =\frac{\sqrt{2}}{2}\,m$

The resultant force is found by Coulomb's law and principle of superposition:

$\vec R = \vec F_{13}+\vec F_{23}$ (1)

Please notice that particles with charges of <em>same</em> sign attract each other and particles with charges of <em>opposite</em> sign repeal each other.

$\vec R = \frac{k\cdot q_{1}\cdot q_{3}}{r_{13}^{2}}\cdot \vec u_{13} +\frac{k\cdot q_{2}\cdot q_{3}}{r_{23}^{2}}\cdot \vec u_{23}$ (2)

Where:

$k$ - Electrostatic constant, in newton-square meters per square Coulomb.
$q_{1}$ , $q_{2}$ , $q_{3}$ - Electric charges, in Coulombs.
$r_{13}$ , $r_{23}$ - Distances between particles, in meters.
$\vec u_{13}$ , $\vec u_{23}$ - Unit vectors, no unit.

If we know that $k = 8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}}$ , $q_{1} = 2\times 10^{-6}\,C$ , $q_{2} = 2\times 10^{-6}\,C$ , $q_{3} = 2\times 10^{-6}\,C$ , $r_{13} =\frac{\sqrt{2}}{2}\,m$ , $r_{23} =\frac{\sqrt{2}}{2}\,m$ , $\vec u_{13} = \left(-\frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{2} \right)$ and $\vec u_{23} = \left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right)$ , then the vector force on charge $q_{3}$ is:

$\vec R = \frac{\left(8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot (2\times 10^{-6}\,C)\cdot (2\times 10^{-6}\,C)}{\left(\frac{\sqrt{2}}{2}\,m \right)^{2}} \cdot \left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right) + \frac{\left(8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot (2\times 10^{-6}\,C)\cdot (2\times 10^{-6}\,C)}{\left(\frac{\sqrt{2}}{2}\,m \right)^{2}} \cdot \left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right)$

$\vec R = 0.072\cdot \left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right) + 0.072\cdot \left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right)\,[N]$

$\vec R = 0.072\cdot \left(0, -\sqrt{2}\right)\,[N]$

And the magnitude of the <em>electrical</em> force on charge $q_{3}$ ( $R$ ), in newtons, due to the others is found by Pythagorean theorem:

$R = 0.102\,N$

The magnitude of <em>electrical</em> force on charge $q_{3}$ due to the others is 0.102 newtons. $\blacksquare$

To learn more on Coulomb's law, we kindly invite to check this verified question: brainly.com/question/506926

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2 years ago