Unit 6: Motion and Forces
... G = the gravitational constant (6.67 x 10-11 N *m2/kg2) M1= Mass of object 1 M2 = Mass of object 2 d = distance between two objects ...
... G = the gravitational constant (6.67 x 10-11 N *m2/kg2) M1= Mass of object 1 M2 = Mass of object 2 d = distance between two objects ...
( ) 13.0m / s ( ( ) 8.0m / s ( ( ) 8m / s ( ( ) 7.2m / s (
... 9.8m / s 2 position 3 the horizontal component of the velocity remains v2 = 15.26m / s . The horizontal distance traveled is simply x = v2t = (15.26m / s ) ( 2.39s ) = 36.5m b) It is noted that the linear speed at position 3 ( v3 = 27.9m / s ) is greater than at position 1 ( v2 = 25m / s ), even tho ...
... 9.8m / s 2 position 3 the horizontal component of the velocity remains v2 = 15.26m / s . The horizontal distance traveled is simply x = v2t = (15.26m / s ) ( 2.39s ) = 36.5m b) It is noted that the linear speed at position 3 ( v3 = 27.9m / s ) is greater than at position 1 ( v2 = 25m / s ), even tho ...
KEY Chapter 8 – Rotational Motion Chapter 6 – Work, Energy
... J = area under the graph = ½ (20.0 N)(4.0 s) = 40.0 N•s J = Δp J = mΔv = 40.0 N•s Δv = (40.0 N•s)/(4.0 kg) = 10.0 m/s 33. a) Calculate the impulse suffered by a 105 kg man who lands on firm ground after jumping from a height of 1.5 m. vf =√ (2ad) = √[(2)(9.8)(1.5)] = 5.4 m/s J = Δp = mΔv = (105)(5.4 ...
... J = area under the graph = ½ (20.0 N)(4.0 s) = 40.0 N•s J = Δp J = mΔv = 40.0 N•s Δv = (40.0 N•s)/(4.0 kg) = 10.0 m/s 33. a) Calculate the impulse suffered by a 105 kg man who lands on firm ground after jumping from a height of 1.5 m. vf =√ (2ad) = √[(2)(9.8)(1.5)] = 5.4 m/s J = Δp = mΔv = (105)(5.4 ...
Year 13 Momentum - Rogue Physicist
... 1) A spherical mass of 1.5kg travelling at a velocity of 2.5m/s collides with another spherical mass of 0.75kg, which is initially at rest. The collision is inelastic such that they join and move away with a common velocity. a) What is the initial momentum of each mass? b) By applying the principal ...
... 1) A spherical mass of 1.5kg travelling at a velocity of 2.5m/s collides with another spherical mass of 0.75kg, which is initially at rest. The collision is inelastic such that they join and move away with a common velocity. a) What is the initial momentum of each mass? b) By applying the principal ...
This worksheet uses the concepts of rotational
... and the condition for rolling without slipping was imposed. Then more generally, the total kinetic energy would be given by: K total 12 Mv2 1 It’s interesting to note that when a wheel rolls without slip, there is one point in contact with the surface which is stationary meaning the friction ...
... and the condition for rolling without slipping was imposed. Then more generally, the total kinetic energy would be given by: K total 12 Mv2 1 It’s interesting to note that when a wheel rolls without slip, there is one point in contact with the surface which is stationary meaning the friction ...
