Plan of Lectures - The Budker Group
... most commonly used systems are SI (MKS) and Gaussian (CGS). Many working physicists still use CGS as it is particular convenient for E&M. In mechanics, it really does not matter, and we will use all kinds of units. Warning: watch out for unit consistency. Use of the K&K book. We will heavily rely on ...
... most commonly used systems are SI (MKS) and Gaussian (CGS). Many working physicists still use CGS as it is particular convenient for E&M. In mechanics, it really does not matter, and we will use all kinds of units. Warning: watch out for unit consistency. Use of the K&K book. We will heavily rely on ...
PHET Forces and Motion Basics Simulator Classwork
... Select the play button on the image. You should now be able to operate the applet. Experiment with the controls until everyone has reached this stage. Your teacher will give you further instruction on how to proceed. Select “Net Force” from the bottom of the screen or main menu, check the boxe ...
... Select the play button on the image. You should now be able to operate the applet. Experiment with the controls until everyone has reached this stage. Your teacher will give you further instruction on how to proceed. Select “Net Force” from the bottom of the screen or main menu, check the boxe ...
uniform circular motion
... and the different angular velocities, . • C) Objects 1 and 2 have different linear velocities, v, and the same angular velocity, . • D) Objects 1 and 2 have different linear velocities, v, and the different angular velocities, . ...
... and the different angular velocities, . • C) Objects 1 and 2 have different linear velocities, v, and the same angular velocity, . • D) Objects 1 and 2 have different linear velocities, v, and the different angular velocities, . ...
Work, Power, and Energy [CH 14
... • Work is the product of force and distance – For a force to do work on an object, some of the force must act in the same direction as the object moves. If there is no movement, no work is done. – Any part of a force that does not act in the direction of motion does no work on an ...
... • Work is the product of force and distance – For a force to do work on an object, some of the force must act in the same direction as the object moves. If there is no movement, no work is done. – Any part of a force that does not act in the direction of motion does no work on an ...
Conservation of Linear Momentum Solutions
... third laws of motion? Explain. Yes. The first law states that an object’s motion is unchanged unless affected by a net external force; same with an object’s momentum. The third law states that when two object’s interact, their affect on one another is equal in magnitude and opposite in direction. Eq ...
... third laws of motion? Explain. Yes. The first law states that an object’s motion is unchanged unless affected by a net external force; same with an object’s momentum. The third law states that when two object’s interact, their affect on one another is equal in magnitude and opposite in direction. Eq ...
Electric potential
... important - not the absolute potential values; electric potential is defined with respect to some arbitrarily chosen zero-point - there is no “absolute zero of potential” usually (but not always): potential is defined in such a way that it is zero at infinity. potential (difference) is also called “ ...
... important - not the absolute potential values; electric potential is defined with respect to some arbitrarily chosen zero-point - there is no “absolute zero of potential” usually (but not always): potential is defined in such a way that it is zero at infinity. potential (difference) is also called “ ...
Review E: Simple Harmonic Motion and Mechanical Energy
... differential equations which states that a unique solution exists which satisfies a given set of initial conditions x0 ≡ x ( t = 0 ) and v0 ≡ v ( t = 0 ) where x0 and v0 are constants. A second approach is to guess the solution and then verify that the guess satisfies the SHO differential equation. ...
... differential equations which states that a unique solution exists which satisfies a given set of initial conditions x0 ≡ x ( t = 0 ) and v0 ≡ v ( t = 0 ) where x0 and v0 are constants. A second approach is to guess the solution and then verify that the guess satisfies the SHO differential equation. ...
Chapter 9 PPT
... the sum of the externally applied forces is zero, and the sum of the externally applied torques is zero. ...
... the sum of the externally applied forces is zero, and the sum of the externally applied torques is zero. ...
CBSE Class 9 Work Energy and Power Quick Study Chapter...
... Total energy E = PE + KE = mg (h-x) + mgx = mgh – mgx + mgx EB = mgh Case - 2: During free fall let the body reaches the position C when fall to A. At C, Potential energy PE = 0 Velocity of the body C is v2 = u2 + 2as u = 0, a = g, s = h ...
... Total energy E = PE + KE = mg (h-x) + mgx = mgh – mgx + mgx EB = mgh Case - 2: During free fall let the body reaches the position C when fall to A. At C, Potential energy PE = 0 Velocity of the body C is v2 = u2 + 2as u = 0, a = g, s = h ...
THE HYDROGEN ATOM (1) Central Force Problem (2) Rigid Rotor
... Separation of Variables: Let ψ (r,θ, φ) = R(r) Yml(θ, φ) Then one can divide through by Yml(θ, φ) to get an eqn. for R, which defines the Central Force Problem: [- (h2/2m) (∂2/∂r2 + (2/r) ∂/∂r ) + l(l+1) h2/ (2m r2) + V(r) - E] R(r) = 0. This is the eqn. for the radial part of the wavefunction. One ...
... Separation of Variables: Let ψ (r,θ, φ) = R(r) Yml(θ, φ) Then one can divide through by Yml(θ, φ) to get an eqn. for R, which defines the Central Force Problem: [- (h2/2m) (∂2/∂r2 + (2/r) ∂/∂r ) + l(l+1) h2/ (2m r2) + V(r) - E] R(r) = 0. This is the eqn. for the radial part of the wavefunction. One ...