Determine the moment of the force f about an axis extending between a and c. express the result as a cartesian vector. a is at the origin, c is at (4,3,0) and f = (4i + 12j -3k) lb coming out of b (4,3,-2).

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).

Carbon 14 is a naturally occurring stable

1.)the isotope cobalt-60 has a nuclear mass of 59.933820 u calculate the mass defect of cobalt-60 using the following information. mass of proton: 1.007825 u mass of neutron: 1.008665 u 1 u = 931.5 mev 2.)the isotope cobalt-60 has a nuclear mass of 59.933820 u calculate the binding energy of cobalt-60 using the following information. mass of proton: 1.007825 u mass of neutron: 1.008665 u 1 u = 931.5 mev 3.)the isotope cobalt-60 has a nuclear mass of 59.933820 u calculate the binding energy per nucleon of cobalt-60 using the following information. mass of proton: 1.007825 u mass of neutron: 1.008665 u 1 u = 931.5 mev