a body of mass 6 kg is floating in a liquid with 2 by 3 of its volume inside the liquid find bouyance force acting on the body​

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).

The radius of an atom is closest in size to a

Aball is thrown from the top of a building with an initial velocity of 21.9 m/s straight upward, at an initial height of 51.6 m above the ground. the ball just misses the edge of the roof on its way down, as shown in the figure. (a) determine the time needed for the ball to reach its maximum height. (b) determine the maximum height. (c) determine the time needed for the ball to return to the height from which it was thrown, and the velocity of the ball at that instant. (d) determine the time needed for the ball to reach the ground. (e) determine the velocity and position of the ball at t = 5.35 s.