Work & Energy
... How much work was done by Fg on the sled? How much work was done by Fup on the sled? If the force of kinetic friction was 20.0 N, how much work was done by friction on the sled? ...
... How much work was done by Fg on the sled? How much work was done by Fup on the sled? If the force of kinetic friction was 20.0 N, how much work was done by friction on the sled? ...
Monday, April 1, 2013
... 4. You must show the detail of your OWN work in order to obtain any credit. Monday, April 1, 2013 ...
... 4. You must show the detail of your OWN work in order to obtain any credit. Monday, April 1, 2013 ...
lab 3: newton`s second law of motion
... time. The term speed does not specify in which direction the object is moving. By contrast, the term velocity not only specifies speed, but also specifies in which direction the object is moving. Velocity is therefore a vector quantity, as explained in chapter 2 of your text, and speed is a scalar q ...
... time. The term speed does not specify in which direction the object is moving. By contrast, the term velocity not only specifies speed, but also specifies in which direction the object is moving. Velocity is therefore a vector quantity, as explained in chapter 2 of your text, and speed is a scalar q ...
THE NOT SO SIMPLE PENDULUM
... frictionless pivot at the point of support. A horizontal rod (1/2 " diameter) mounted with a clamp on a platform attached to the wall provided the rigid support. The force probe has a mounting hole in its body at one end and it is hung on the horizontal rod through this hole. The friction between th ...
... frictionless pivot at the point of support. A horizontal rod (1/2 " diameter) mounted with a clamp on a platform attached to the wall provided the rigid support. The force probe has a mounting hole in its body at one end and it is hung on the horizontal rod through this hole. The friction between th ...
Topics 1, 2, 3, 4, 9 selected problems paper 1 take
... A truck collides head on with a less massive car moving in the opposite direction to the truck. During the collision, the average force exerted by the truck on the car is FT and the average force exerted by the car on the truck is FC. Which one of the following statements is correct? A. ...
... A truck collides head on with a less massive car moving in the opposite direction to the truck. During the collision, the average force exerted by the truck on the car is FT and the average force exerted by the car on the truck is FC. Which one of the following statements is correct? A. ...
Work/Power/Energy Lab
... 1. Each group of students should consist of two students: student A, and student B. Both students will perform the experiment in the stairway at the end of the hallway, toward the exit way. a. Both students measure their mass in kilograms and calculate their weight – force in Newton’s. b. If the mas ...
... 1. Each group of students should consist of two students: student A, and student B. Both students will perform the experiment in the stairway at the end of the hallway, toward the exit way. a. Both students measure their mass in kilograms and calculate their weight – force in Newton’s. b. If the mas ...
Forces and Motion Commotion 2012
... Part A: Motion and Speed of Objects 1. Describe and measure motion using the concept of a reference point. 2. Describe and measure speed and be able to calculate speed. (Know what 2 factors –distance and time—on which speed depends.) 3. Graph motion showing changes in distance as a function of time ...
... Part A: Motion and Speed of Objects 1. Describe and measure motion using the concept of a reference point. 2. Describe and measure speed and be able to calculate speed. (Know what 2 factors –distance and time—on which speed depends.) 3. Graph motion showing changes in distance as a function of time ...
m/s 2 - mrhsluniewskiscience
... 1. Find net force (by combining vectors). 2. Calculate acceleration (using 2nd law). ...
... 1. Find net force (by combining vectors). 2. Calculate acceleration (using 2nd law). ...
Work, Power & Energy
... Select the equation and solve: KE = ½ m v2 KE = ½ (0.145 kg)(35.8 m/s)2 KE = ½ (0.145 kg)(1281.54 m/s/s) KE = ½ (185.8 kg m/s/s) KE = 92.9 kg m/s/s, or 92.9 Nm, or 92.9J ...
... Select the equation and solve: KE = ½ m v2 KE = ½ (0.145 kg)(35.8 m/s)2 KE = ½ (0.145 kg)(1281.54 m/s/s) KE = ½ (185.8 kg m/s/s) KE = 92.9 kg m/s/s, or 92.9 Nm, or 92.9J ...
work-energy
... Select the equation and solve: KE = ½ m v2 KE = ½ (0.145 kg)(35.8 m/s)2 KE = ½ (0.145 kg)(1281.54 m/s/s) KE = ½ (185.8 kg m/s/s) KE = 92.9 kg m/s/s, or 92.9 Nm, or 92.9J ...
... Select the equation and solve: KE = ½ m v2 KE = ½ (0.145 kg)(35.8 m/s)2 KE = ½ (0.145 kg)(1281.54 m/s/s) KE = ½ (185.8 kg m/s/s) KE = 92.9 kg m/s/s, or 92.9 Nm, or 92.9J ...
wbm-physics
... Energy seems to be stored in some form related to height. This energy is related to the position of a body, not its motion. It is called potential energy. – measures potential or possibility for work to be done. (Some kinds of potential energy are related to things other than height.) ...
... Energy seems to be stored in some form related to height. This energy is related to the position of a body, not its motion. It is called potential energy. – measures potential or possibility for work to be done. (Some kinds of potential energy are related to things other than height.) ...
Conservation of Energy name: Conservation of energy while rolling
... Where h is the height of the hill and vf is the final velocity at the bottom of the hill. Note that if mechanical energy was conserved, and no mechanical energy was converted to thermal energy by friction, then this ratio would always be equal to one. If mechanical energy is lost due to friction, th ...
... Where h is the height of the hill and vf is the final velocity at the bottom of the hill. Note that if mechanical energy was conserved, and no mechanical energy was converted to thermal energy by friction, then this ratio would always be equal to one. If mechanical energy is lost due to friction, th ...