All the 5`s - The Physics Teacher
... (h) The work function of tungsten is 4.50 eV. Calculate the maximum kinetic energy of an electron ejected from a tungsten surface when electromagnetic radiation whose photon energy is 5.85 eV shines on the surface. Energy of incident photon = Work function + KE of electron 1.35 (eV) / 2.16 × 10−19 J ...
... (h) The work function of tungsten is 4.50 eV. Calculate the maximum kinetic energy of an electron ejected from a tungsten surface when electromagnetic radiation whose photon energy is 5.85 eV shines on the surface. Energy of incident photon = Work function + KE of electron 1.35 (eV) / 2.16 × 10−19 J ...
Mechanics 2
... To build on the work in Mechanics 1 by extending the range of mechanics concepts which students are able to use in modelling situations. Students will be able to use the rigid body model in simple cases involving moments. Students will be expected to formulate models, using the mechanics within this ...
... To build on the work in Mechanics 1 by extending the range of mechanics concepts which students are able to use in modelling situations. Students will be able to use the rigid body model in simple cases involving moments. Students will be expected to formulate models, using the mechanics within this ...
Figure 1: Problem 1 Figure 2: Problem 2 1. The spring is unstretched
... where m2 and m3 are the mass of the 2and 3-kg cylinders, y20 and y30 are the initial coordinates of the 2- and 3-kg cylinders. It has the relationship between (y2 − y20 ) and (y3 − y30 ) as y2 − y20 = −2(y3 − y30 ). We apply the work-energy equation to ...
... where m2 and m3 are the mass of the 2and 3-kg cylinders, y20 and y30 are the initial coordinates of the 2- and 3-kg cylinders. It has the relationship between (y2 − y20 ) and (y3 − y30 ) as y2 − y20 = −2(y3 − y30 ). We apply the work-energy equation to ...
Electromagnetic Theory
... In the above integral we have the delta function δ ( t – t′ – x – X ( t′ ) ⁄ c ) = δ ( t′ + x – X ( t′ ) ⁄ c – t ) so that g ( t′ ) = t′ + x – X ( t′ ) ⁄ c – t Differentiating this with respect to t′ : dg ( t′ ) ∂ x – X ( t′ ) --------------- = ġ ( t′ ) = 1 + ------------------------dt′ ∂ t′ c To d ...
... In the above integral we have the delta function δ ( t – t′ – x – X ( t′ ) ⁄ c ) = δ ( t′ + x – X ( t′ ) ⁄ c – t ) so that g ( t′ ) = t′ + x – X ( t′ ) ⁄ c – t Differentiating this with respect to t′ : dg ( t′ ) ∂ x – X ( t′ ) --------------- = ġ ( t′ ) = 1 + ------------------------dt′ ∂ t′ c To d ...
6.8 Mb - Todd Satogata
... G = 6.67384 × 10−11 m3 kg−1 s−2 § This formula for the general gravitational force was developed by Newton (1687) § It’s a much much better approximation, but again only an approximation! § Superceded by Einstein’s general theory of relativity (1916) § But this equation works well enough to ...
... G = 6.67384 × 10−11 m3 kg−1 s−2 § This formula for the general gravitational force was developed by Newton (1687) § It’s a much much better approximation, but again only an approximation! § Superceded by Einstein’s general theory of relativity (1916) § But this equation works well enough to ...
Student AP Physics 1 Date Oscillations – MC 1. A mass m, attached
... 22. Which of the following statements about energy is correct? (A) The potential energy of the spring is at a minimum at x = 0 (B) The potential energy of the spring is at a minimum at x = A (C) The kinetic energy of the block is at a minimum at x =0 (D) The kinetic energy of the block is at a maxim ...
... 22. Which of the following statements about energy is correct? (A) The potential energy of the spring is at a minimum at x = 0 (B) The potential energy of the spring is at a minimum at x = A (C) The kinetic energy of the block is at a minimum at x =0 (D) The kinetic energy of the block is at a maxim ...
