In the diagram, 91 = +6.60 x 10-6 C, 92 = +3.10 x 10-6 C, and 43 = +5.30 x 10-6 C.
Find the electric potential energy of 92.
(Remember, energy is NOT a vector.)
0.350 m -
92
91
0.155 m
93
(Unit = J)

Aplane flies directly between two cities, a and b, which are separated by . from a to b, the plane flies into a headwind. on the return trip from b to a, the wind velocity is unchanged. the trip from b to a takes less than the trip from a to b. what is the airspeed of the plane, assuming it is the same in both directions

Classify the following soils according to the uscs classification system. provide appropriate uscs designations. a) this sample of well-graded gravel with sand was obtained from a large earthen dam in vancouver, canada. the sample has 73% fine to coarse sub-angular gravel, 25% fine to coarse sub-angular sand and 2% fines. the maximum size of the particles is 75 mm. the coefficient of curvature is 2.7, while the uniformity coefficient is 12.4. b) this dark brown and wet soil with a “strong organic odor” has 100% passing the no. 200 sieve. the liquid limit of the material is 32% when not dried and is 21% when oven-dried. the plastic index is 21% when not dried. c) this sand has 61% predominantly fine sand, 23% silty fines, and 16% fine sub-rounded gravel size. the maximum size is 20 mm. the liquid limit is 33% and the plastic limit is 27%. d) this soil has 74% fine to coarse sub-angular reddish sand and 26% organic and silty dark brown fines. the liquid limit is 37% when not dried and is 26% when oven dried. the plastic index is 6% when not dried.

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).