4.30 m long string vibrates at 640 Hz with successive nodes at 1.90m, 2.50 m and 3.10 m from one end of the string.

A1000-kg weather rocket is launched straight up. the rocket motor provides a constant acceleration for 16-s, then the motor stops. the rocket altitude 20-s after launch is 510-m. you can ignore any effects of air resistance. a. what was the rocket’s acceleration during the first 16-s? b. what is the rocket’s speed as it passes through a cloud 5100-m above the ground?

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 law of reflection states that the angle of reflection is equal to the angle of