T3F2008
... C. In the figure below, two blocks, of masses m1 and m2, are connected by a massless cord that is wrapped around a uniform disk of rotational inertia, I and radius R. The disk can rotate without friction about a fixed horizontal axis through its center; the cord cannot slip on the disk. The system i ...
... C. In the figure below, two blocks, of masses m1 and m2, are connected by a massless cord that is wrapped around a uniform disk of rotational inertia, I and radius R. The disk can rotate without friction about a fixed horizontal axis through its center; the cord cannot slip on the disk. The system i ...
Kinematics Vf = vi + at D = vit + ½ a t = vi + 2ad V = d/t Speed versus
... Newton’s 1 st Law a) A body at rest tends to stay at rest tends to stay at rest, unless acted upon by an outside force b) a body in motion tends to stay in motion, in a straight line at constant speed, unless acted upon by an outside force Newton’s 2 nd Law F=ma Newton’s 3 rd Law For ever ...
... Newton’s 1 st Law a) A body at rest tends to stay at rest tends to stay at rest, unless acted upon by an outside force b) a body in motion tends to stay in motion, in a straight line at constant speed, unless acted upon by an outside force Newton’s 2 nd Law F=ma Newton’s 3 rd Law For ever ...
7. What is the momentum of an 8 kg bowling ball rolling at 2 m/s
... 4. Two carts with masses 4.0 kg and 3.0 kg respectively, move on a frictionless track with velocities of 5.0 m/s and 4.0 m/s. The carts stick together after colliding head-on. Find the final speed of the two carts. (1.14 m/s) ...
... 4. Two carts with masses 4.0 kg and 3.0 kg respectively, move on a frictionless track with velocities of 5.0 m/s and 4.0 m/s. The carts stick together after colliding head-on. Find the final speed of the two carts. (1.14 m/s) ...
Potential Energy
... • Make sure there are only conservative forces and kinetic energy in the problem • Identify all the potential and kinetic energy at the beginning • Identify all the potential and kinetic energy at the end • Set the initial and final energy equal to one another ...
... • Make sure there are only conservative forces and kinetic energy in the problem • Identify all the potential and kinetic energy at the beginning • Identify all the potential and kinetic energy at the end • Set the initial and final energy equal to one another ...
PowerPoint Presentation - ABOUT TEAL
... Draw (imaginary) line parallel to the force. If distance from axis measured perpendicular to this line (lever arm) is clear, then the torque is the force times this distance ...
... Draw (imaginary) line parallel to the force. If distance from axis measured perpendicular to this line (lever arm) is clear, then the torque is the force times this distance ...
force-problems-old
... 6. A 70 kg man slides down a rope that serves as a fire escape. The maximum force that can be applied to the rope without it breaking is 600 N. Find the minimum acceleration the man can have without breaking the rope? Find the maximum acceleration the man can have? 7. A 50 kg child slides down the s ...
... 6. A 70 kg man slides down a rope that serves as a fire escape. The maximum force that can be applied to the rope without it breaking is 600 N. Find the minimum acceleration the man can have without breaking the rope? Find the maximum acceleration the man can have? 7. A 50 kg child slides down the s ...
Law of Conservation of Momentum
... pTOTAL (i ) pTOTAL ( f ) Momentum can be transferred from one part of a system to another, but the total amount remains constant. Momentum is conserved during any interaction among objects – collisions, explosions, etc. – since these interactions involve INTERNAL impulses (one part of the system ...
... pTOTAL (i ) pTOTAL ( f ) Momentum can be transferred from one part of a system to another, but the total amount remains constant. Momentum is conserved during any interaction among objects – collisions, explosions, etc. – since these interactions involve INTERNAL impulses (one part of the system ...
For Physics - Career Point Kota
... (B) the pressure exerted by the water vapour enclosed in the cooker decreases the boiling point of water (C) the water vapour pressure does not influence the boiling point of the water (D) there is no loss of heat to the surrounding as the pressure cooker is airtight. ...
... (B) the pressure exerted by the water vapour enclosed in the cooker decreases the boiling point of water (C) the water vapour pressure does not influence the boiling point of the water (D) there is no loss of heat to the surrounding as the pressure cooker is airtight. ...
Quiz 6
... At the equilibrium position, the net torque acting on the loop is zero. Choosing rotational axis along the hinged side, we have Στ = 0 ⇒ NIBL²cosθ-mg(L/2)sinθ = 0 ⇒ tanθ = 2NIBL/(mg) ...
... At the equilibrium position, the net torque acting on the loop is zero. Choosing rotational axis along the hinged side, we have Στ = 0 ⇒ NIBL²cosθ-mg(L/2)sinθ = 0 ⇒ tanθ = 2NIBL/(mg) ...
Newton`s 2nd Law
... In the equation for weight, g is no longer considered the acceleration due to gravity, but rather the gravitational field strength, with units of newtons/kilogram. Inertial and gravitational masses have been tested and are believed to always be equal in amount. This is why all objects freefall at th ...
... In the equation for weight, g is no longer considered the acceleration due to gravity, but rather the gravitational field strength, with units of newtons/kilogram. Inertial and gravitational masses have been tested and are believed to always be equal in amount. This is why all objects freefall at th ...
