Question

Notify Me Bookmark

A particle of charge Q is fixed at the origin of an xy coordinate system. At t = 0 a particle (m = 0.910 g, q = 5.53 μC) is located on the x axis at x = 19.2 cm, moving with a speed of 26.7 m/s in the positive y direction. For what value of Q will the moving particle execute circular motion? (Neglect the gravitational force on the particle.) Number Units

A particle of charge Q is fixed at the origin of an xy coordinate system. At t = 0 a particle (m = 0.910 g, q = 5.53 μC) is located on the x axis at x = 19.2 cm, moving with a speed of 26.7 m/s in the positive y direction. For what value of Q will the moving particle execute circular motion? (Neglect the gravitational force on the particle.) Number Units

Image text
A particle of charge Q is fixed at the origin of an x y coordinate system. At t = 0 a particle ( m = 0.910 g , q = 5.53 μ C ) is located on the x axis at x = 19.2 c m , moving with a speed of 26.7 m / s in the positive y direction. For what value of Q will the moving particle execute circular motion? (Neglect the gravitational force on the particle.)
Number i Units ◻

Detailed Answer

0 0

Please enter your email here. You will get email when this question is answered by our tutor.

Doubtrix Logo

Problem is not yet solved!

Click on Notify me to get email when question is answered.

Join Doubtrix with 7-days free trial

  • Icon Correct & Verified Answers
  • Icon No AI-Generated Answers
  • Icon Video Solution for Difficult Questions
  • Icon Work With Experts to Reach at Correct Answers
Notify me Login Sign Up

Similar/Related Questions

Identical 45 μC charges are fixed on an x axis at x = ±3.1 m. A particle of charge q = −13 μC is then released from rest at a point on the positive part of the y axis. Due to the symmetry of the situation, the particle moves along the y axis and has kinetic energy 1.1 J as it passes through the point x = 0, y = 4.2 m. (a) What is the kinetic energy of the particle as it passes through the origin? (b) At what negative value of y will the particle momentarily stop? (a) Number Units (b) Number Units
Solution
0 0

Two particles are fixed on an x axis. Particle 1 of charge 40.8 μC is located at x = −11.1 cm; particle 2 of charge Q is located at x = 11.2 cm. Particle 3 of charge magnitude 42.8 μC is released from rest on the y axis at y = 11.1 cm. What is the value of Q if the initial acceleration of particle 3 is in the positive direction of (a) the x axis and (b) the y axis? (a) Number Units (b) Number Units
Solution
0 0

Two particles are located in an xy coordinate system. Particle 1 (mass m1 = 10.0 kg) is fixed at the origin. Particle 2 (mass m2 = 20.0 kg) is initially located on the x axis (θ = 0) at distance r = 2.00 m from the origin and particle 1. This situation is depicted in the simulation (linked below). Simulation The Gravitational Force Question 1 For the given position of the two particles, express the gravitational force on particle 1 due to particle 2, F→12, in unit-vector notation. Glive your answer in nanonewtons, and fill in the boxes below. F→12 = (nN)i^+(nN)j^ Click if you would like to Show Work for this question: Open Show Worik Attempts: 0 of 8 used SAVE FOR LATER SUBMIT ANSWER Question 2 For the default position of particle 2, express the gravitational force on particle 2 due to particle 1, F→21, in unit-vector notation. Give your answer in nanonewitons, and fill in the boxes below. F→21 = (nN)i^+(nN)j^ Question 4 What value for the angle θ in the range 0 ≤ θ ≤ 90∘ results in the following gravitational force on particle 1:F→12 = (22.0 nN)i^+(8.88 nN)j^? θ = the tolerance is +/−2% Click if you would like to Show Work for this question: Open Show Work Question 5 What value for the distance r between the two particles results in the following gravitational force on particle 1: F→12 = (22.0 nN)i^+(8.88 nN)j^? r = m the tolerance is +/−2%Two particles are located in an xy coordinate system. Particle 1 (mass m1 = 10.0 kg) is fixed at the origin. Particle 2 (mass m2 = 20.0 kg) is initially located on the x axis (θ = 0) at distance r = 2.00 m from the origin and particle 1. This situation is depicted in the simulation (linked below). Simulation The Gravitational Force Question 1 For the given position of the two particles, express the gravitational force on particle 1 due to particle 2, F→12, in unit-vector notation. Glive your answer in nanonewtons, and fill in the boxes below. F→12 = (nN)i^+(nN)j^ Click if you would like to Show Work for this question: Open Show Worik Attempts: 0 of 8 used SAVE FOR LATER SUBMIT ANSWER Question 2 For the default position of particle 2, express the gravitational force on particle 2 due to particle 1, F→21, in unit-vector notation. Give your answer in nanonewitons, and fill in the boxes below. F→21 = (nN)i^+(nN)j^ Question 4 What value for the angle θ in the range 0 ≤ θ ≤ 90∘ results in the following gravitational force on particle 1:F→12 = (22.0 nN)i^+(8.88 nN)j^? θ = the tolerance is +/−2% Click if you would like to Show Work for this question: Open Show Work Question 5 What value for the distance r between the two particles results in the following gravitational force on particle 1: F→12 = (22.0 nN)i^+(8.88 nN)j^? r = m the tolerance is +/−2%
Solution
0 0