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solong [7]
2 years ago
13

Four charges are at the corners of a square centered at the origin as follows q at a a 2q at a a 3q at a a and 6q at a a A fifth

charge q is placed at the origin and released from rest Find its speed when it is a great distance from the origin if a 1.5 m q 2.8 mu or micro CC and its mass is 0.4 kg

Physics
1 answer:
Wewaii [24]2 years ago
7 0

Answer:

Vfx = 0.42 √x  and Vfy = - 0.729 √x

Explanation:

To solve this problem let's use the coulomb law applied to each load

        F = k q1 q2 / r²

Let's look for the distance from a corner to the center of the square

        r² = (a / 2) ² + (a / 2)²

        r = √ (a / 2)² 2

        r = a / 2 √2

Let's write the forces, in the attached there is a diagram of the charges

Charge q in the corner

         F1 = k q q / r²

         F1 = k q² / r²

         F1 = k q² 2 / a²

         F1 = 8.99 10⁹ (2.8 10⁻⁶)²   2 / 1.5²

         F1 = 62.65 10⁻³ N = 6.265 10⁻² N

charge 2q in another corner

        F2 = k 2q q / r²

        F2 = 2 k q² / r²

Notice that this force is twice the force F1, since the distances are equal

        F2 = 2 F1

3q charge

        F3 = k 3q q / r²

        F3 = 3 F1

6q charge

        F4 = k 6q q / r²

        F4 = 6 F1

Let's calculate the total force, for this it is useful to break down each force into its components and then add them, let's use trigonometry

        cos θ = Fx / F

        Fx = F cos θ

        sin θ = Fy / F

        Fy = F sin θ

Let's add each force

X axis

         Fx = F1x - F2x - F3x + F4x

         Fx = F1 cos 45 - 2F1 cos 45 - 3F1 cos 45 + 6 F1 cos 45

         Fx = 2 F1 cos 45

         Fx = 2 (6,265 10-2) cos 45

         Fx = 8.86 10-2 N

Axis y

         Fy = F1y + F2y -F3y -F4y

         Fy = F1 sin 45 + 2 F1 sin 45 - 3 F1 sin 45 - 6 F1 sin  45

         Fy = - 6Fi sin 45

         Fy = -6 (6,265 10⁻²) sin45

         Fy = - 26.54 10⁻² N

As we have the accelerations, we can find the speed on each axis

X axis

         Vf² = Vo² + 2aₓ x

         Vfx² = 2aₓ x

         Vfx² = 2 8.86 10⁻² x

         Vfx = 0.42 √x

Axis y

         Vfy² = vfy² + 2ay Y

         Vfy² = 2 ay Y

         Vfy² = -2 26.54 10-2 x

         Vfy = - 0.729 √x

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A bicycle tire is spinning counterclockwise at 2.70 rad/s. During a time period Δt = 1.50 s, the tire is stopped and spun in the
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Answer:

Δω = -5.4 rad/s

αav = -3.6 rad/s²

Explanation:

<u>Given</u>:

           Initial angular velocity = ωi = 2.70 rad/s

           Final angular velocity = ωf = -2.70 rad/s (negative sign is  

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           Change in time period = Δt = 1.50 s

<u>Required</u>:

           Change in angular velocity = Δω = ?

           Average angular acceleration = αav = ?

<u>Solution</u>:

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               Δω = ωf - ωi

               Δω = -2.70 - 2.70

               Δω = -5.4 rad/s.

          <u> Average angular acceleration (αav):</u>

               αav = Δω/Δt

               αav = -5.4/1.50

              αav = -3.6 rad/s²

Since, the angular velocity is decreasing from 2.70 rad/s (in counter clockwise direction) to rest and then to -2.70 rad/s (in clockwise direction) so, the change in angular velocity is negative.

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Answer:43.34 m

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After this object will start moving under gravity

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After fuel run out distance traveled in upward direction is

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