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2 years ago

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A 5. 0-c charge is 10 m from a small test charge. what is the magnitude of the force experienced by a 1. 0 nc charge placed at t

he location of the test charge?

1 answer:

sveta [45]2 years ago

4 0

The magnitude of the force experienced by a 1. 0 c charge placed at the location of the test charge is 4.5*10^8N.

To find the answer, we have to know more about the force experienced between two charges.

<h3>How to find the force experienced between two charges?</h3>

It is given that,

$Q_1=5C\\Q_2=1C\\r=10m.$

We have to find the force experienced between two charges,

$F=\frac{kQ_1Q_2}{r^2}$

where, k=8.99*10^9.

Thus, by substitution, the force will be,

$F=\frac{8.99*10^9*5*1}{10^2} =4.5*10^8N$

Thus, we can conclude that, the magnitude of the force experienced by a 1. 0 c charge placed at the location of the test charge is 4.5*10^8N.

Learn more about the force here:

brainly.com/question/4086258

#SPJ4

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A 1.65 kg mass stretches a vertical spring 0.260 m If the spring is stretched an additional 0.130 m and released, how long does

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Answer:

The system will take approximately 0.255 seconds to reach the (new) equilibrium position.

Explanation:

We notice that block-spring system depicts a Simple Harmonic Motion, whose equation of motion is:

$y(t) = A\cdot \cos \left(\sqrt{\frac{k}{m} }\cdot t +\phi\right)$ (1)

Where:

$y(t)$ - Position of the mass as a function of time, measured in meters.

$A$ - Amplitude, measured in meters.

$k$ - Spring constant, measured in newtons per meter.

$m$ - Mass of the block, measured in kilograms.

$t$ - Time, measured in seconds.

$\phi$ - Phase, measured in radians.

The spring is now calculated by Hooke's Law, that is:

$k = \frac{m\cdot g}{\Delta y}$ (2)

Where:

$g$ - Gravitational acceleration, measured in meters per square second.

$\Delta y$ - Deformation of the spring due to gravity, measured in meters.

If we know that $m=1.65\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ and $\Delta y = 0.260\,m$ , then the spring constant is:

$k = \frac{(1.65\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{0.260\,m}$

$k = 62.237\,\frac{N}{m}$

If we know that $A = 0.130\,m$ , $k = 62.237\,\frac{N}{m}$ , $m=1.65\,kg$ , $x(t) = 0\,m$ and $\phi = 0\,rad$ , then (1) is reduced into this form:

$0.130\cdot \cos (6.142\cdot t)=0$ (1)

And now we solve for $t$ . Given that cosine is a periodic function, we are only interested in the least value of $t$ such that mass reaches equilibrium position. Then:

$\cos (6.142\cdot t) = 0$

$6.142\cdot t = \cos^{-1} 0$

$t = \frac{1}{6.142}\cdot \left(\frac{\pi}{2} \right)\,s$

$t \approx 0.255\,s$

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Answer:

The mass of the block m is:

$m=M(sin(\theta)+\mu_{s}cos(\theta))$

Explanation:

Let's analyze the block by parts

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Where:

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