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Alekssandra [29.7K]
3 years ago
11

Two charged particles, with charges q1=q and q2=4q, are located at a distance d= 2.00cm apart on the x axis. A third charged par

ticle, with charge q3=q, is placed on the x-axis such that the magnitude of the force that charge 1 exerts on charge 3 is equal to the force that charge 2 exerts on charge 3. Find the position of charge 3 when q = 2.00 nC . Assuming charge 1 is located at the origin of the x-axis and the positive x-axis points to the right, find the two possible values x3,r, and x3,l for the position of charge 3.
Physics
1 answer:
Murrr4er [49]3 years ago
7 0

Answer:

X₃₁ = 0.58 m  and  X₃₂ = -1.38 m

Explanation:

For this exercise we use Newton's second law where the force is the Coulomb force

        F₁₃ - F₂₃ = 0

        F₁₃ = F₂₃

Since all charges are of the same sign, forces are repulsive

        F₁₃ = k q₁ q₃ / r₁₃²

        F₂₃ = k q₂ q₃ / r₂₃²

Let's find the distances

         r₁₃ = x₃- 0

         r₂₃ = 2 –x₃

We substitute

      k q q / x₃² = k 4q q / (2-x₃)²

      q² (2 - x₃)² = 4 q² x₃²

        4- 4x₃ + x₃² = 4 x₃²

        5x₃² + 4 x₃ - 4 = 0

We solve the quadratic equation

        x₃ = [-4 ±√(16 - 4 5 (-4)) ] / 2  5

        x₃ = [-4 ± 9.80] 10

       X₃₁ = 0.58 m

       X₃₂ = -1.38 m

For this two distance it is given that the two forces are equal

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The graph shows the federal budget from 1980 to 2010. In which period did the federal budget show the greatest deficit?
kodGreya [7K]

Answer:

2000 to 2010

Explanation:

sorry I'm just guessing

8 0
2 years ago
6. The hole on a level, elevated golf green is a horizontal distance of 150 m from the tee and at an elevation of 12.4 m above t
Georgia [21]

Answer:

u = 104.68 m/s

Explanation:

given,

horizontal distance = 150 m

elevation of  12.4 m

angle = 8.6°

horizontal motion = x = u cos θ. t .............(1)

vertical motion =

y = u sin \theta - \dfrac{1}{2}gt^2................(2)

from equation(1) and (2)

y = x tan \theta - \dfrac{gx^2}{2u^2cos^2\theta}..........{3}

12.4 = 150\times tan (8.6) - \dfrac{9.8\times 150^2}{2u^2cos^2(8.6)}

\dfrac{9.8\times 150^2}{2u^2cos^2(8.6)} = 10.29

\dfrac{9.8\times 150^2}{2\times 10.29\times cos^2(8.6)} = u^2

u = \sqrt{10959.34}

u = 104.68 m/s

The initial speed of the ball is u = 104.68 m/s

8 0
3 years ago
1. A train engine pulls the boxcars. What are the horizontal forces present?(1 point)
NikAS [45]

Force is the influence that is capable of bringing about motion

The correct options are;

1. C. The pull of the engine, friction force

2. A. Weight, push of the tracks against the wheel

3. D. The astronaut weighs the most on Earth and least on the Moon

4. Please see attached drawing for example

5. Please see attached drawing for example

6. The data points will fall along a line this that as the force increases, the acceleration increases

Reason:

1. The horizontal forces are the forces acting along the horizontal x-axis, therefore;

A train engine pulling a boxcar what horizontal forces are;

C. The pull of the engine, friction force

2. The vertical forces are the forces acting an  the direction of the y-axis of a graph

The vertical forces present when a train pulls a boxcar are;

A. Weight, push of the tracks against the wheel

3. The weight of an astronaut, W = mass × Gravity

  • Therefore, the weight increases with increasing acceleration due to gravity

Therefore, the least weight is on the moon, where acceleration due to gravity is 1.6 m/s², and most on Earth where the acceleration due to gravity is approximately 9.81 m/s²

  • The correct option is D. The astronaut weighs the most on Earth and least on the Moon

4. The magnitude of the downward force is the sum of the vertical forces acting downwards

5. The net force is given by the vector sum of the forces

6. According to Newton's second law, Force, F = Mass, m × Acceleration, a

  • Therefore, where the students conduct the experiments rightly, we have that the data points will fall along a line this that as the force increases, the acceleration increases

Learn more here;;

brainly.com/question/19657850

5 0
2 years ago
A constant friction force of 25 N acts on a 65-kg skier for 15 s on level snow. What is the skier’s change in velocity
nikdorinn [45]

Answer:

\Delta v=5.77m/s

Explanation:

Newton's 2nd Law relates the net force <em>F</em> on an object of mass <em>m </em>with the acceleration <em>a</em> it experiments by <em>F=ma.</em> In our case the net force is the friction force, since it's the only one the skier is experimenting horizontally and the vertical ones cancel out since he's not moving in that direction. Our acceleration then will be:

a=\frac{F}{m}

Also, acceleration is defined by the change of velocity \Delta v in a given time t, so we have:

a=\frac{\Delta v}{t}

Since we want the change in velocity, <em>mixing both equations</em> we conclude that:

\Delta v=at=\frac{Ft}{m}

Which for our values means:

\Delta v=\frac{Ft}{m}=\frac{(25N)(15s)}{(65Kg)}=5.77m/s

4 0
4 years ago
How does the number 10 relate to light, sound, force and motion?
Oksana_A [137]

Answer:

number 10 is the value of acceleration due to gravity.

8 0
4 years ago
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