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Mariulka [41]
2 years ago
8

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?
Physics
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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3 years ago
A projectile fired up into the air at an angle has a range of 235 m and a flight time of 47 s.
madam [21]
<h2>Answer:5ms^{-1},133.6m,51.18ms^{-1}</h2>

Explanation:

Let v_{x},v_{y} be the horizontal and vertical components of velocity.

Question a:

Horizontal component of velocity is the ratio of range and time of flight.

So,horizontal component of velocity is \frac{235}{47}=5ms^{-1}

So,v_{x}=5ms^{-1}

Question b:

Time of flight=\frac{2v_{y}}{g}

So,v_{y}=\frac{47\times 9.8}{2}=51.18ms^{-1}

Maximum height is given by \frac{v_{y}^{2}}{2g}

So,maximum height is \frac{51.18^{2}}{2\times 9.8}=133.6m

Question c:

The vertical velocity is already calculated in Question b.

v_{y}=51.18ms^{-1}

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3 years ago
A quantity found by multiplying the force by the distance moved
Andrej [43]
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What is the energy of an electromagnetic wave that has a frequency of
sukhopar [10]

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Explanation:

Given the following data;

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Planck's constant, h = 6.626 x 10-34 J·s.

To find the energy of the electromagnetic wave;

Mathematically, the energy of an electromagnetic wave is given by the formula;

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Where;

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h represents Planck's constant.

f is the frequency of a wave.

Substituting the values into the formula, we have;

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