Two ice skaters push off against one another starting from a stationary position. The 45.0-kg skater acquires a speed of 0.375 m/s. What speed does the 60.0-kg skater acquire in m/s

Two red blood cells each have a mass of 9.0 multiply.gif 10-14 kg and carry a negative charge spread uniformly over their surfaces. the repulsion arising from the excess charge prevents the cells from clumping together. one cell carries −2.50 pc of charge and the other −3.10 pc, and each cell can be modeled as a sphere 7.5 μm in diameter. (a) what speed would they need when very far away from each other to get close enough to just touch? assume that there is no viscous drag from any of the surrounding liquid. answer is: 321 m/s how do i find this (b) what is the maximum acceleration of the cells in part (a)? answer is: 1.38e+10 m/s2 how did they find this answer

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).

Dinate system, and a charge q = - 2.00 nc is placed s2 10 n> c. (a) what is the change in electric potential energy when the dipole moment of a molecule changes its orientation with respect to es from parallel to perpendicular? (b) at what absolute temperature t is the average translational kinetic energy 32 kt of a molecule equal to the change in potential energy calculated in part(: abovethistemperature,thermalagitationprevents the dipoles from aligning with the electric field.) placed in a uniform electric field e with magnitude 1.6 * 6 on the positive x-axis at x = 4.00 cm. (a) if a third charge q3 = +6.00 nc is now placed at the point x = 4.00 cm, y = 3.00 cm, find the x- and y-components of the total force exerted on this charge by the other two. (b) find the magnitude and direction of figure p21.62 ll uu 21.55 . torque on a dipole. an electric dipole with dipole moment ps is in a uniform external electric field es. (a) find the orientations of the dipole for which the torque on the dipole is zero. (b) which of the orientations in part (a) is stable, and which is unstable? (hint: consider a small rotation away from the equi- librium position and see what happens.) (c) show that for the stable orientation in part (b), the dipole’s own electric field tends to oppose the external field. 21.62 ..