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pishuonlain [190]
2 years ago
13

Two charged particles are a distance of 1.82 m from each other. One of the particles has a charge of 7.03 nC, and the other has

a charge of 4.14 nC.
Physics
1 answer:
Naya [18.7K]2 years ago
3 0

The magnitude of the electric force is(F) 79.07\times 10^{-9} N and the electric force between the charges is repulsive.

<h3>How can we calculate the magnitude of the electric force?</h3>

To calculate the magnitude of the electric force that one particle exerts on the other,  we are using the formula,

F=k\frac{q_{1} q_{2} }{r^{2} }

Here we are given,

q_1= Charge of the first particle= 7.03nC = 7.03\times 10^{-9}  C.

q_2=Charge of the second particle = 4.14nC= 4.14\times 10^{-9} C.

r= The distance between two charges=1.82m.

k= Constant of proportionality= \frac{1}{4\pi \epsilon}=9\times 10^{9} Nm^{2} /C^{2}

Now we put the known values in the above equation, we get

F=k\frac{q_{1} q_{2} }{r^{2} }

Or,F=9\times 10^{9} \times \frac{7.03\times 10^{-9}\times 4.14\times 10^{-9} }{(1.82)^{2} }

Or, F= 79.07\times 10^{-9} N.

From the above calculation we can conclude that the magnitude of the electric force(F) is 79.07\times 10^{-9} N.

Charges q_1 and q_2 is 79.07\times 10^{-9} C and 4.14\times 10^{-9}C respectively, Both charges are positive and have same polarity. The force between the same charges is repulsive.

Thus, we can conclude that, the electric force between the charges is repulsive.

Learn more about Charged particle:

brainly.com/question/18763828

#SPJ4

Disclaimer: The question was given incomplete in the portal. Here is the complete question.

Question: Two charged particles are a distance of 1.82 m from each other. One of the particles has a charge of 7.03 nC, and the other has a charge of 4.14 nC.

  • What is the magnitude of the electric force that one particle exerts on the other?
  • Is the force attractive or repulsive?
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A running mountain lion can make a leap 10.0 m long, reaching a maximum height of 3.0 m.?a.What is the speed of the mountain lio
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Explanation:

This is going to be long, so if you want to see how it was solved refer to the attached solution. If you want to know the step by step process, read on.

To solve this, you will need use two kinematic equations and SOHCAHTOA:

d = v_it + \dfrac{1}{2}at^{2}\\\\vf = vi + at

With these formulas, we can derive formulas for everything you need:

Things you need to remember:

  • A projectile at an angle has a x-component (horizontal movement) and y-component (vertical movement), which is the reason why it creates an angle.
  • Treat them separately.
  • At maximum height, the vertical final velocity is always 0 m/s going up. And initial vertical velocity is 0 m/s going down.
  • Horizontal movement is not influenced by gravity.
  • acceleration due to gravity (a) on Earth is constant at 9.8m/s

First we need to take your given:

10.0 m long (horizontal) and maximum height of 3.0m (vertical).

d_x=10.0m\\d_y=3.0m

What your problem is looking for is the initial velocity and the angle it left the ground.

Vi = ?     Θ =?

Vi here is the diagonal movement and do solve this, we need both the horizontal velocity and the vertical velocity.

Let's deal with the vertical components first:

We can use the second kinematic equation given to solve for the vertical initial velocity but we are missing time. So we use the first kinematic equation to derive a formula for time.

d_y=V_i_yt+\dfrac{1}{2}at^{2}

Since it is at maximum height at this point, we can assume that the lion is already making its way down so the initial vertical velocity would be 0 m/s. So we can reduce the formula:

d_y=0+\dfrac{1}{2}at^{2}

d_y=\dfrac{1}{2}at^{2}

From here we can derive the formula of time:

t=\sqrt{\dfrac{2d_y}{a}}

Now we just plug in what we know:

t=\sqrt{\dfrac{(2)(3.0m}{9.8m/s^2}}\\t=0.782s

Now that we know the time it takes to get from the highest point to the ground. The time going up is equal to the time going down, so we can use this time to solve for the intial scenario of going up.

vf_y=vi_y+at

Remember that going up the vertical final velocity is 0m/s, and remember that gravity is always moving downwards so it is negative.

0m/s=vi_y+-9.8m/s^{2}(0.782s)\\-vi_y=-9.8m/s^{2}(0.782s)\\-vi_y=-7.66m/s\\vi_y=7.66m/s

So we have our first initial vertical velocity:

Viy = 7.66m/s

Next we solve for the horizontal velocity. We use the same kinematic formula but replace it with x components. Remember that gravity has no influence horizontally so a = 0:

d_x=V_i_xt+\dfrac{1}{2}0m/s^{2}(t^{2})\\d_x=V_i_xt

But horizontally, it considers the time of flight, from the time it was released and the time it hits the ground. Also, like mentioned earlier the time going up is the same as going down, so if we combine them the total time in flight will be twice the time.

T= 2t

T = 2 (0.782s)

<em>T = 1.564s</em>

<em>So we use this in our formula:</em>

<em>d_x=V_i_xT\\\\10.0m=Vi_x(1.564s)\\\\\dfrac{10.0m}{1.564s}=V_i_x\\\\6.39m/s=V_i_x</em>

Vix=6.39m/s

Now we have the horizontal and the vertical component, we can solve for the diagonal initial velocity, or the velocity the mountain lion leapt and the angle, by creating a right triangles, using vectors (see attached)

To get the diagonal, you just use the Pythagorean theorem:

c²=a²+b²

Using it in the context of our problem:

Vi^{2}=Viy^2+Vix^2\\Vi^2=(7.66m/s)^2+(6.39m/s)^2\\\sqrt{Vi}=\sqrt{(7.66m/s)^2+(6.39m/s)^2}\\\\Vi=9.98m/s

The lion leapt at 9.98m/s

Using SOHCAHTOA, we know that we can TOA to solve for the angle, because we have the opposite and adjacent side:

Tan\theta=\dfrac{O}{A}\\\\Tan\theta=\dfrac{V_i_y}{V_i_x}\\\\\theta=Tan^{-1}\dfrac{V_i_y}{V_i_x}\\\\\theta=Tan^{-1}\dfrac{7.66m/s}{6.39m/s}\\\\\theta=50.17

The lion leapt at an angle of 50.16°.

6 0
2 years ago
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