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Unformatted text preview: far away is the cave wall? Use for the speed of sound, 340 m/s. Sol 3. The emitted sound wave travels twice the distance between the bat and the wall, in hitting the wall and echoing back to the bat. We assume that the speed at which the bat is ying is much smaller than the speed of sound, so that the position of the bat has not varied appreciably with respect to the wall during these 0.2 s. Under this assumption, the distance to the cave wall is given by d = 1 2 (340m/s . 2s) = 34m 4. By what distance must two charges of +4 C be separated so that the repulsive force between them is 3 . 6 10 10 N. (Coulombs constant, k = 9 10 9 Nm 2 / C 2 .) Sol 4. We use the Coulomb force law F = k q 1 q 2 r 2 r = kq 1 q 2 F = (9 10 9 Nm 2 C-2 ) (+4C) (+4C) 3 . 6 10 10 N = 2m 1...
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- Fall '08