
L33
... • Thus far we have been dealing only with what is called geometrical optics • In geometrical optics we deal only with the behavior of light rays it either travels in a straight line or is reflected by a mirror, or bent (refracted) when it travels from one medium into another. • However, light is a ...
... • Thus far we have been dealing only with what is called geometrical optics • In geometrical optics we deal only with the behavior of light rays it either travels in a straight line or is reflected by a mirror, or bent (refracted) when it travels from one medium into another. • However, light is a ...
1 - CNU.edu
... elbow joint as the axis of rotation. [1.79] N m (b) If the net torque obtained in part (a) is nonzero, in which direction will the forearm and hand rotate? [clockwise] (c) Would the net torque exerted on the forearm and hand? Why? ...
... elbow joint as the axis of rotation. [1.79] N m (b) If the net torque obtained in part (a) is nonzero, in which direction will the forearm and hand rotate? [clockwise] (c) Would the net torque exerted on the forearm and hand? Why? ...
Today: Quantum mechanics
... In a test of eye sensitivity, experimenters used 1 millisecond (0.001 s) flashes of green light. The lowest power light that could be seen was 4x10-14 Watt. How many green (500 nm, 2.5 eV) photons is this? A. 10 photons B. 100 photons C. 1,000 photons D. 10,000 photons Tues. Nov. 17, 2009 ...
... In a test of eye sensitivity, experimenters used 1 millisecond (0.001 s) flashes of green light. The lowest power light that could be seen was 4x10-14 Watt. How many green (500 nm, 2.5 eV) photons is this? A. 10 photons B. 100 photons C. 1,000 photons D. 10,000 photons Tues. Nov. 17, 2009 ...
This worksheet uses the concepts of rotational
... rotating about its axis, the rotational kinetic energy is given by: K rot 12 I2 . If, in addition, the center of mass of the wheel is translating, then the translational kinetic energy is given by K trans 12 Mv 2 . To find the total kinetic energy of the system, add the individual contributions ...
... rotating about its axis, the rotational kinetic energy is given by: K rot 12 I2 . If, in addition, the center of mass of the wheel is translating, then the translational kinetic energy is given by K trans 12 Mv 2 . To find the total kinetic energy of the system, add the individual contributions ...
Momentum and Impulse
... The same change in momentum may be the result of a SMALL force exerted for a LONG time, or a LARGE force exerted for a SHORT time. ...
... The same change in momentum may be the result of a SMALL force exerted for a LONG time, or a LARGE force exerted for a SHORT time. ...
Chapter 3 Problem Set
... Before we can solve for power we have to convert the time (25 min) into seconds: t = 25 min X 60 sec/min = 1,500 sec Now solving for the power: P = W/t = 253,820 J/1500 sec = 169 w (watts) 24. A boy throws a 4-kg pumpkin at 8 m/sec to a 40-kg girl on roller skates, who catches it. At what speed does ...
... Before we can solve for power we have to convert the time (25 min) into seconds: t = 25 min X 60 sec/min = 1,500 sec Now solving for the power: P = W/t = 253,820 J/1500 sec = 169 w (watts) 24. A boy throws a 4-kg pumpkin at 8 m/sec to a 40-kg girl on roller skates, who catches it. At what speed does ...
Angular Momentum
... vi = 2 m/s in a circle of radius ri = 0.2 m. The cord is then slowly pulled from below, decreasing the radius of the circle to r = 0.1 m. What is the puck’s speed at the smaller radius? Find the tension in the cord at the smaller radius. May 22, 2017 ...
... vi = 2 m/s in a circle of radius ri = 0.2 m. The cord is then slowly pulled from below, decreasing the radius of the circle to r = 0.1 m. What is the puck’s speed at the smaller radius? Find the tension in the cord at the smaller radius. May 22, 2017 ...
Document
... 13. Loss in PE by Y = gain in PE of X + gain in KE by X plus Y Gain in KE of the system = 5 (10) (2) – 4(10)(2 sin 30o) = 60 J 14. From A to B, friction acts down the incline. Work done by friction = Fs = 2.6 x AB cos 180o = -2.6 x 10/sin 30o = - 52 J From B to C, friction acts up the incline . Work ...
... 13. Loss in PE by Y = gain in PE of X + gain in KE by X plus Y Gain in KE of the system = 5 (10) (2) – 4(10)(2 sin 30o) = 60 J 14. From A to B, friction acts down the incline. Work done by friction = Fs = 2.6 x AB cos 180o = -2.6 x 10/sin 30o = - 52 J From B to C, friction acts up the incline . Work ...
chapter 9 notes physics 2
... rotation is placed. You typically will choose the location so that the lines of action of one or more of the unknown forces pass through the axis. This will simplify the torque equation making the math easier. Remember the direction of your force indicates whether it is positive or negative with res ...
... rotation is placed. You typically will choose the location so that the lines of action of one or more of the unknown forces pass through the axis. This will simplify the torque equation making the math easier. Remember the direction of your force indicates whether it is positive or negative with res ...
Class: XII Subject: Physics Topic: Electromagnetic Waves No. of
... 14. A radiation of energy E falls normally on a perfectly reflecting surface. The momentum transferred to the surface is a. E/c b. 2 E/c c. Ec d. E/c2 ...
... 14. A radiation of energy E falls normally on a perfectly reflecting surface. The momentum transferred to the surface is a. E/c b. 2 E/c c. Ec d. E/c2 ...
Experiment 9 - WFU Physics
... very likely will fall to some lower orbit. The energy can be emitted in the form of light. This light is visible provided the energy lost by the atom lies in the range of about 2.8 10-19 J to 5.0 10-19 J for each electronic transition. (This is approximately the range of energies per atom which is ...
... very likely will fall to some lower orbit. The energy can be emitted in the form of light. This light is visible provided the energy lost by the atom lies in the range of about 2.8 10-19 J to 5.0 10-19 J for each electronic transition. (This is approximately the range of energies per atom which is ...
photoeffect
... » There was a threshold frequency for ejection Classical physics failed to explain this, Lenard won the Nobel Prize in Physics in 1905. ...
... » There was a threshold frequency for ejection Classical physics failed to explain this, Lenard won the Nobel Prize in Physics in 1905. ...