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Newtons laws
Newtons laws

... • When the high points of each surface touch they bond, you must break these bonds to move one of the pieces (origin of static friction) • As surfaces move across each other, electrostatic forces continue to attract between high points, resulting in the weaker kinetic friction • Air drag (any fluid) ...
’ m = 22.0 kg       µ
’ m = 22.0 kg µ

... Thus, in uniform circular motion there must be a net force to produce the centripetal acceleration. The centripetal force is the name given to the net force required to keep an object moving on a circular path. The direction of the centripetal force always points toward the center of the circle and ...
lecture14
lecture14

... Figure out all forces and their points of application Sum all forces and divide by mass to find COM’s linear acceleration For each force, compute perp-dot-product from COM to point of force application and add value into total torque of COM Divide total torque by the MOI at the COM to find angular a ...
Discussion Examples Chapter 5: Newton`s Laws of Motion
Discussion Examples Chapter 5: Newton`s Laws of Motion

AP Physics IB
AP Physics IB

... If the net force is zero . . . • The object is not moving or . . . • The object is moving at a constant velocity therefore . . . • The object is in equilibrium ...
Circular Motion
Circular Motion

... center of the circle is called the centripetal force To analyze centripetal acceleration situations accurately, you must identify the agent of the force that causes the acceleration (such as tension on a string). Then you can apply Newton’s second law for the component in the direction of the accele ...
Document
Document

Circular Motion
Circular Motion

Equilibrium and the Equilibrant
Equilibrium and the Equilibrant

Newton`s Laws of Motion
Newton`s Laws of Motion

... interchangeably because the only comparison we have is the Earth’s gravity. • Weight will change based on local gravity; NASA has to take this into effect ...
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r - God and Science

... Thus, in uniform circular motion there must be a net force to produce the centripetal acceleration. The centripetal force is the name given to the net force required to keep an object moving on a circular path. The direction of the centripetal force always points toward the center of the circle and ...
ch05
ch05

... Thus, in uniform circular motion there must be a net force to produce the centripetal acceleration. The centripetal force is the name given to the net force required to keep an object moving on a circular path. The direction of the centripetal force always points toward the center of the circle and ...
ert146 lect kinetic of motion
ert146 lect kinetic of motion

... SI system: In the SI system of units, mass is a base unit and weight is a derived unit. Typically, mass is specified in kilograms (kg), and weight is calculated from W = mg. If the gravitational acceleration (g) is specified in units of m/s2, then the weight is expressed in newtons (N). On the earth ...
香港考試局
香港考試局

Torque
Torque

HW4 - due 6 pm Day 8 (Wed. Aug. 6)
HW4 - due 6 pm Day 8 (Wed. Aug. 6)

Causes of circular motion
Causes of circular motion

Newton`s Second Law (F=ma)
Newton`s Second Law (F=ma)

... _____ 8. A 10-kg object has greater acceleration due to gravity than a 5-kg object. _____ 9. The acceleration of an object equals its mass times the net force applied to it. _____ 10. The acceleration of an object due to gravity depends on the object’s initial velocity ...
(the terminal velocity is smaller for larger cross
(the terminal velocity is smaller for larger cross

... Force exerted by a spring: Hooke’s law: If spring is stretched or compressed by some small amount it exerted a force which is linearly proportional to the amount of stretching or compressing. The constant of proportionality is called the spring constant ...
Physics Homework
Physics Homework

IPC Force Momentum Freefall Newtons Law Test Review
IPC Force Momentum Freefall Newtons Law Test Review

AP Physics B
AP Physics B

... A pilot executes a vertical dive, then follows a semi-circular arc until it is going straight up. Just as the plane is at its lowest point, the force on him is a. less than mg, and pointing up. b. less than mg, and pointing down. c. more than mg, and pointing up. d. more than mg, and pointing down. ...
θ ω ω - Assignment Expert
θ ω ω - Assignment Expert

香港考試局
香港考試局

... which is free to rotate about the other end O as shown. The object is swung to rotate in a vertical circle so that it attains an angular speed of 6 rad s-1 at its topmost position. What is the force exerted on one end of the rod at this instant ? A. a compressive force of 7.6 N B. a tensional force ...
Topic 6 Problem Set 2016
Topic 6 Problem Set 2016

... Two masses of 4.501022 kg each are located along the x-axis. The first mass is at the origin, and the second mass is at x = +1.25106 m. 44. Find the gravitational field strength at the point on the x-axis directly between the two masses. 45. Find the gravitational field strength at x = +2.50106 m ...
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Jerk (physics)

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