1, 3, 6, 10, 11, 17, 21 / 1, 4, 12, 15, 20, 24, 28, 36, 38
... The force of air resistance will always act in the direction that is opposite to the direction of motion of the ball. The net force on the ball is the resultant of the weight and the force of air resistance. a. As the ball moves upward, the force of air resistance acts downward. Since air resistance ...
... The force of air resistance will always act in the direction that is opposite to the direction of motion of the ball. The net force on the ball is the resultant of the weight and the force of air resistance. a. As the ball moves upward, the force of air resistance acts downward. Since air resistance ...
Ph211_CH5_worksheet-f06
... Since the masses are attached their accelerations are equal: a1y = a2x = asystem Solving for asystem: m2gsin – m1asystem - m1g = m2asystem asystem = (m2gsin – m1g)/(m1 + m2) = -1.03 m/s2 (i.e. up the incline!) e. What are the tension forces acting on each mass? Express the tension vectors in compo ...
... Since the masses are attached their accelerations are equal: a1y = a2x = asystem Solving for asystem: m2gsin – m1asystem - m1g = m2asystem asystem = (m2gsin – m1g)/(m1 + m2) = -1.03 m/s2 (i.e. up the incline!) e. What are the tension forces acting on each mass? Express the tension vectors in compo ...
Impulse, momentum, and center of mass
... Impulse equals change in momentum, so the left-hand magnet will change momentum by +F·Δt the right-hand magnet with change momentum by -F·Δt. Clearly, the total change in momentum for the system is zero, and this derivation works regardless of whether or not the magnets began at rest. So internal fo ...
... Impulse equals change in momentum, so the left-hand magnet will change momentum by +F·Δt the right-hand magnet with change momentum by -F·Δt. Clearly, the total change in momentum for the system is zero, and this derivation works regardless of whether or not the magnets began at rest. So internal fo ...
PH202 Chapter 14 solutions
... attached to a spring of negligible mass and force constant . The block is free to move on a frictionless horizontal surface, while the left end of the spring is held fixed. When the spring is neither compressed nor stretched, the block is in equilibrium. If the spring is stretched, the block is disp ...
... attached to a spring of negligible mass and force constant . The block is free to move on a frictionless horizontal surface, while the left end of the spring is held fixed. When the spring is neither compressed nor stretched, the block is in equilibrium. If the spring is stretched, the block is disp ...
1 Newton`s First and Second Laws
... will change. It also states that the larger the object’s mass, the smaller its acceleration will be. Acceleration is the change in motion of an object. Think about a baseball. If you throw the baseball gently, it will not move very fast. If you throw it more forcefully, it will move much more quickl ...
... will change. It also states that the larger the object’s mass, the smaller its acceleration will be. Acceleration is the change in motion of an object. Think about a baseball. If you throw the baseball gently, it will not move very fast. If you throw it more forcefully, it will move much more quickl ...
Forces and acceleration Newton`s 2nd Law
... ** Position the photogate so that it's right next to the leading edge of the card. The photogate should be in gate mode. Press the reset button to reset the time to zero. ** Slowly ease the cart forward (not letting it go) so that it starts the photogate (red light on) and then back the cart up just ...
... ** Position the photogate so that it's right next to the leading edge of the card. The photogate should be in gate mode. Press the reset button to reset the time to zero. ** Slowly ease the cart forward (not letting it go) so that it starts the photogate (red light on) and then back the cart up just ...
Physics 513 Name Vaughan Worksheet Newton`s Second Law
... 21. A hockey puck (with a mass of 0.5 kg) is sliding across the ice with an initial velocity of 4 m/s East. It slows down and comes to rest over 100 meters. a) What is the magnitude and direction of the frictional force? b) What is the coefficient of friction? 22. Find the coefficient of kinetic fr ...
... 21. A hockey puck (with a mass of 0.5 kg) is sliding across the ice with an initial velocity of 4 m/s East. It slows down and comes to rest over 100 meters. a) What is the magnitude and direction of the frictional force? b) What is the coefficient of friction? 22. Find the coefficient of kinetic fr ...
Multiple choice test template
... 8) A lorry weighing 3 tonnes is travelling at 10 ms-1. The force needed to stop it in 10 seconds is (a) 306 N (c) 29400 N (e) I don’t know ...
... 8) A lorry weighing 3 tonnes is travelling at 10 ms-1. The force needed to stop it in 10 seconds is (a) 306 N (c) 29400 N (e) I don’t know ...