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- can you please answer (a) & (b)?Solution: The projectile motion equations: x = x₁ + (vo cos 0) t y = yo + (vo sin 0o) t - 1/2gt² Call the starting point (xo, yo) = (0,0). To find (x, y) for the ending point, use triangle trig on the hill angled at 20°: x = d cos 20° = .940d and y = d sin 20° = .342d Plug in: .940d = (vo coso) t (Eqn. 1) .342d = (vo sin 0o) tgt² (Eqn. 2) The variable t is not mentioned in the problem. We can remove it by solving Eqn. 1 for t and plugging into Eqn. 2: t = .940d vo cos 00 .940d vo cos o -) - 12/19 ( Cleaning up and plugging in g = 9.8m: .342d = (v₁ sino) ( .940d vo cos 00 2Problem 1: A block having a mass of m= 14 kg is suspended via two cables as shown in the figure. The angles shown in the figure are as follows: a = 12° and B= 28°. Cable l Cable 2- m
- The angle between the vectors that connect ther carbon atom to the hydrogen atoms in the methane molecule can be calculated using the dot product or the cross product rule. In the case of the dot product, we get the correct answer, which is 109.5 degree, but if we use the cross product, the answer might be 70.5 degree. Why?intersecting cylinders defined by x + y2 6 to and x? + z? = a' is V= 16a/3. 1-36. Find the value of the integral fs A da, where A = xi - yj + zk and Sis the closed surface defined by the cylinder c x + y2. The top and bottom of the cylinder are at z = d and 0, respectively.Ctrl (a) Find the average rate of change of the area of a circle with respect to its radius r as r changes from 2 to each of the following. (1) 2 to 3 A Tab (ii) 2 to 2.5 (iii) 2 to 2.1 Need Help? (b) Find the instantaneous rate of change when r = 2. A'(2) = 11 Submit Answer 1 Esc Shift Caps Lock X 1 X Type here to search Q X Read It A X F2 2 Watch It F3 # 3 O F4 Alt View Previous Question Question 4 of 6 View Next Question S D $ 4 WE R E Z X C F5 D F % 5 F6 T ( 6 G 0 & 7 H FB U * 8 FO F10 9 70°F Clear A V B N M J1 K 2 L 0 Alt Activa Go to Se P F12