Day 1 Notes: Dealing with projectiles in two dimensions. There are
... The theoratic and experimental results were too different from each other. The percent error was approximately 75 pecent. This difference in results came from errors resulting from human’s reaction time when we pressed the stop watch and also the inability to spin the mass and the string in constan ...
... The theoratic and experimental results were too different from each other. The percent error was approximately 75 pecent. This difference in results came from errors resulting from human’s reaction time when we pressed the stop watch and also the inability to spin the mass and the string in constan ...
Motion Characteristics for Circular Motion
... also can be used to guide our thinking about the variables in the equation and how they relate to each other. For instance, the equation suggests that for objects moving around circles of different radii in the same period, the object traversing the circle of a larger radius must be traveling with t ...
... also can be used to guide our thinking about the variables in the equation and how they relate to each other. For instance, the equation suggests that for objects moving around circles of different radii in the same period, the object traversing the circle of a larger radius must be traveling with t ...
Motion of a Point Charge in a Magnetic Field
... We have already seen the basics of this material in our laboratory on the e/m of the electron. There we learned that the magnetic force on a point charge moving with a velocity in a G G G magnetic field is given by FB = qv × B . If the charge has its initial velocity perpendicular to the magnetic fi ...
... We have already seen the basics of this material in our laboratory on the e/m of the electron. There we learned that the magnetic force on a point charge moving with a velocity in a G G G magnetic field is given by FB = qv × B . If the charge has its initial velocity perpendicular to the magnetic fi ...
PHY 131–002 - Oakton Community College
... 10) On a frictionless track, car A, with mass mA = 0.300 kg, moves to the right toward car B with velocity vA = +2.00 m/s. Car B, with mass mB = 0.250 kg, moves toward car A with a velocity of vB = –1.50 m/s. After the collision, car B has a velocity of v’B = +1.50 m/s. a) What is the velocity of ca ...
... 10) On a frictionless track, car A, with mass mA = 0.300 kg, moves to the right toward car B with velocity vA = +2.00 m/s. Car B, with mass mB = 0.250 kg, moves toward car A with a velocity of vB = –1.50 m/s. After the collision, car B has a velocity of v’B = +1.50 m/s. a) What is the velocity of ca ...
Section Study Guide
... other planets each move in a perfect circle around the sun. b. According to Kepler’s First Law, Earth and the other planets each move in an elliptical orbit with the sun at one focus. 2. t1 = t2 3. T2 r3 4. Newton derived Kepler’s laws from the universal law of gravitation. 5. T = 3.17 107 s; ...
... other planets each move in a perfect circle around the sun. b. According to Kepler’s First Law, Earth and the other planets each move in an elliptical orbit with the sun at one focus. 2. t1 = t2 3. T2 r3 4. Newton derived Kepler’s laws from the universal law of gravitation. 5. T = 3.17 107 s; ...
Linking Asteroids and Meteorites through Reflectance
... Newton’s Universal Law of Gravitation • Every mass attracts every other mass through the force called gravity • Newton came up with this formula F = G M1 M2 r2 M1, M2 are the masses of the two objects r is the distance between the objects G = constant = 6.67 x 10-11 m3/(kgs2) ...
... Newton’s Universal Law of Gravitation • Every mass attracts every other mass through the force called gravity • Newton came up with this formula F = G M1 M2 r2 M1, M2 are the masses of the two objects r is the distance between the objects G = constant = 6.67 x 10-11 m3/(kgs2) ...
Unit 8 Worksheet 1
... c. When the batter hits the ball, a net force of 1320 N, opposite to the direction of the ball's initial motion, acts on the ball for 9.0 x 10-3 s during the hit. What is the change in momentum of the ball? What is the final velocity of the ball? ...
... c. When the batter hits the ball, a net force of 1320 N, opposite to the direction of the ball's initial motion, acts on the ball for 9.0 x 10-3 s during the hit. What is the change in momentum of the ball? What is the final velocity of the ball? ...
A ball on the end of a string is swung in a vertical circle
... moving in a circle at constant speed. Which statement best explains why the speed of the satellite does not change even though there is a net force exerted on it? a) The satellite is in equilibrium. b) The acceleration of the satellite is 0 m/s2. c) The centripetal force has magnitude mv2/r. d) The ...
... moving in a circle at constant speed. Which statement best explains why the speed of the satellite does not change even though there is a net force exerted on it? a) The satellite is in equilibrium. b) The acceleration of the satellite is 0 m/s2. c) The centripetal force has magnitude mv2/r. d) The ...
