An object is thrown directly up (positive direction) with a velocity (v o ) of 20.0 m/s and d o = 0. how high does it rise (v = 0 cm/s at top of rise). remember, acceleration is -9.80 m/s^2 .

Not all tiles will be usedidentify the element present in each star based on the gaps observed in wavelengths of its light refer to this table containing absorption wavelengths (in nm) of several different elements the table ishelium 447, 502 ,587 ,668carbon 427 ,515, 600 ,678sulfur 425, 565, 639, 675xenon 484, 590, 687calcium 429 ,527, 593, 645silver 421 ,521, 662

When a neutral charge, comes into contact with a positive charge they repel each other nothing happens the neutral charge becomes negative they attract each other

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).