The internal energy of a system increases by 36 joules. the system does work using 17 joules. what amount of heat was added to this system? joules\

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).

6–48 bananas are to be cooled from 24 to 138c at a rate of 215 kg/h by a refrigeration system. the power input to the refrigerator is 1.4 kw. determine the rate of cooling, in kj/ min, and the cop of the refrigerator. the specific heat of banana above freezing is 3.35 kj/kg·8c.

Aweight lifter raises a 1600 n barbell to a height of 2.0 meters. how much work was done? w = fd a) 30 joules b) 3000 joules c) 320 joules d) 3200 joules