PHY2049 Spring 2010 Profs. P. Avery, A. Rinzler, S. Hershfield
... Solution: The net amount of charge on both spheres does not change because charge is conserved. Thus, in the end q1 + q2 = 10Q. Also, in the end they have the same potential so kq1 /R = kq2 /(3R), which implies that q2 = 3q1 . Solve these two equations to find in the end q1 = 2.5Q, and q2 = 7.5Q. Th ...
... Solution: The net amount of charge on both spheres does not change because charge is conserved. Thus, in the end q1 + q2 = 10Q. Also, in the end they have the same potential so kq1 /R = kq2 /(3R), which implies that q2 = 3q1 . Solve these two equations to find in the end q1 = 2.5Q, and q2 = 7.5Q. Th ...
Momentum, Impulse, and Collision Review Name: Fill in the
... 13) Using the picture below, rank from least to greatest the ball’s momentum at each point. The ball is rolling down and along the curved slope. ...
... 13) Using the picture below, rank from least to greatest the ball’s momentum at each point. The ball is rolling down and along the curved slope. ...
Section 1
... field. In the magnetic field, other objects can be drawn to the magnet. b. In magnetism, 'to repel' means to experience a force that tends to push them away from each other. If two same forces ( N-N ) are brought near each other they will push away. c. In magnetism, 'to attract' means to experience ...
... field. In the magnetic field, other objects can be drawn to the magnet. b. In magnetism, 'to repel' means to experience a force that tends to push them away from each other. If two same forces ( N-N ) are brought near each other they will push away. c. In magnetism, 'to attract' means to experience ...
Introduction
... manner that their direction at any point is the same as the direction of the electric field at that point. The direction of the field line at any point is the same as the direction in which a positive charge would move it placed at that point. ...
... manner that their direction at any point is the same as the direction of the electric field at that point. The direction of the field line at any point is the same as the direction in which a positive charge would move it placed at that point. ...
Powerpoint Slide
... I (point mass) = Mr2 Moments of inertia add, so our total I for the two masses is: I = I1 + I2 = Mr2 + Mr2 = 2Mr2 What is the effect on I if a triple r? r 3r I = I1 + I2 = M(3r)2 + M(3r)2 = 9Mr2 + 9Mr2 = 18 Mr2 We increased I by a factor of 9!! ...
... I (point mass) = Mr2 Moments of inertia add, so our total I for the two masses is: I = I1 + I2 = Mr2 + Mr2 = 2Mr2 What is the effect on I if a triple r? r 3r I = I1 + I2 = M(3r)2 + M(3r)2 = 9Mr2 + 9Mr2 = 18 Mr2 We increased I by a factor of 9!! ...
Student Text, pp. 378-381
... even causing sparks. One variety of static eliminator uses a radioactive source, polonium-210, that emits positively charged particles (a particles). Explain how this can help to reduce dust on the film. ...
... even causing sparks. One variety of static eliminator uses a radioactive source, polonium-210, that emits positively charged particles (a particles). Explain how this can help to reduce dust on the film. ...
2-27 Potential Energy, Potential, and Work
... 1m apart. The right-hand charge is released. Find its velocity when it is 10cm farther away. E Field and Force are not the same everywhere so Fnet = ma requiresv calculus. =0 v=? Also need a system to handle direction. ...
... 1m apart. The right-hand charge is released. Find its velocity when it is 10cm farther away. E Field and Force are not the same everywhere so Fnet = ma requiresv calculus. =0 v=? Also need a system to handle direction. ...
1988E1. The isolated conducting solid sphere of radius a shown
... In the diagram above, point C is a distance a from the center of the rod (i.e., on the rod's surface), and point D is a distance 3a from the center on a radius that is 90° from point C. Determine the following. i. The potential difference Vc - VD between points C and D ii. The work required by an ex ...
... In the diagram above, point C is a distance a from the center of the rod (i.e., on the rod's surface), and point D is a distance 3a from the center on a radius that is 90° from point C. Determine the following. i. The potential difference Vc - VD between points C and D ii. The work required by an ex ...
The Magnetic Field
... terms of the magnetic force FB that the field exerts on a charged particle moving with a velocity v. • Experiments on charged particles moving in a magnetic field give the following results: – The magnitude FB of the magnetic force exerted on the particle is proportional to the magnitude of the char ...
... terms of the magnetic force FB that the field exerts on a charged particle moving with a velocity v. • Experiments on charged particles moving in a magnetic field give the following results: – The magnitude FB of the magnetic force exerted on the particle is proportional to the magnitude of the char ...
Study on Internal Mechanisms of Charge, Current, Electric Field and
... the impulse u of the electronic angular momentum. According to the traditional theory, the electric field is generated by the charge, and the photon is the medium particle of the electric field. It follows that the electric field force is the force produced by the collision of electronic angular mom ...
... the impulse u of the electronic angular momentum. According to the traditional theory, the electric field is generated by the charge, and the photon is the medium particle of the electric field. It follows that the electric field force is the force produced by the collision of electronic angular mom ...
Classical field theory
... depends entirely on the values of δqi (t) (and possibly its derivatives) taken at the boundary of the integration domain. This can easily be shown by performing a series of integrations by parts. The Euler-Lagrange equations thus govern the dynamics inside the integration domain. This is not influen ...
... depends entirely on the values of δqi (t) (and possibly its derivatives) taken at the boundary of the integration domain. This can easily be shown by performing a series of integrations by parts. The Euler-Lagrange equations thus govern the dynamics inside the integration domain. This is not influen ...
