The capacitor is now reconnected to the battery, and the plate separation is restored to d. A dielectric plate is slowly moved into the capacitor until the entire space between the plates is filled. Find the energy U2 of the dielectric-filled capacitor. The capacitor remains connected to the battery. The dielectric constant is K.

Express your answer in terms of A, d, V, K, and ϵ0.

Moon effect. some people believe that the moon controls their activities. if the moon moves from being directly on the opposite side of earth from you to being directly overhead, by what percentage does (a) the moon's gravitational pull on you increase and (b) your weight (as measured on a scale) decrease? assume that the earth–moon (center-to-center) distance is 3.82 × 10^8 m, earth's radius is 6.37 × 10^6 m, moon's mass is 7.36 × 10^22 kg, and earth's mass is 5.98 × 10^24 kg.

Did the proton move into a region of higher potential or lower potential? did the proton move into a region of higher potential or lower potential? because the proton is a negative charge and it accelerates as it travels, it must be moving from a region of higher potential to a region of lower potential.because the proton is a negative charge and it accelerates as it travels, it must be moving from a region of lower potential to a region of higher potential.because the proton is a positive charge and it slows down as it travels, it must be moving from a region of higher potential to a region of lower potential.because the proton is a positive charge and it slows down as it travels, it must be moving from a region of lower potential to a region of higher potential.request answerpart bwhat was the potential difference that stopped the proton? express your answer with the appropriate units.î”v î” v = nothingnothingrequest answerpart cwhat was the initial kinetic energy of the proton, in electron volts? express your answer in electron volts.ki k i = nothing ev request answerprovide feedback

Part f - example: finding two forces (part i) two dimensional dynamics often involves solving for two unknown quantities in two separate equations describing the total force. the block in (figure 1) has a mass m=10kg and is being pulled by a force f on a table with coefficient of static friction îľs=0.3. four forces act on it: the applied force f (directed î¸=30â above the horizontal). the force of gravity fg=mg (directly down, where g=9.8m/s2). the normal force n (directly up). the force of static friction fs (directly left, opposing any potential motion). if we want to find the size of the force necessary to just barely overcome static friction (in which case fs=îľsn), we use the condition that the sum of the forces in both directions must be 0. using some basic trigonometry, we can write this condition out for the forces in both the horizontal and vertical directions, respectively, as: fcosî¸â’îľsn=0 fsinî¸+nâ’mg=0 in order to find the magnitude of force f, we have to solve a system of two equations with both f and the normal force n unknown. use the methods we have learned to find an expression for f in terms of m, g, î¸, and îľs (no n).