8ps Home Care

E26-37 Use τ = pE sin θ, where θ is the angle between p and E. For this dipole p = qd = 2ed or p = 2(1.6 × 10−19 C)(0.78 × 10−9 m) = 2.5 × 10−28 C · m. For all three cases pE = (2.5 × 10−28 C · m)(3.4 × 106 N/C) = 8.5 × 10−22 N · m. The only thing we care about is the angle. (a) For the parallel case θ = 0, so sin θ = 0, and τ = 0. (b) For the perpendicular case θ = 90◦ , so sin θ = 1, and τ = 8.5 × 10−22 N · m.. (c) For the anti-parallel case θ = 180◦ , so sin θ = 0, and τ = 0. E26-38…

security relative to the model is then constructed. An objective is to design a cryptographic method based on the weakest assumptions possible anticipating a powerful adversary. Asymptotic analysis and usually also worst-case analysis is used and so care must be exercised to determine when proofs have practical significance. In contrast, polynomial attacks which are feasible under the model might, in practice, still be computationally infeasible. Security analysis of this type, although…

x < N then f (x) < M , x3 + 5 < M , x < (M − 5)1/3 , N = (M − 5)1/3 . 67. if δ ≤ 2 then |x − 3| < 2, −2 < x − 3 < 2, 1 < x < 5, and |x2 − 9| = |x + 3||x − 3| < 8|x − 3| < |x − 3| < 1 , δ = min 2, 1 8 8 68. (a) (b) if We don’t care about the value of f at x = a, because the limit is only concerned with values of x near a. The condition that f be defined for all x (except possibly x = a) is necessary, because if some points were…

inconsistent with observations of the physical system it purports to describe, this suggests that either errors have been made in solving the mathematical problem, or the mathematical model itself needs refinement, or observations must be made with greater care. The examples in this section are typical of applications in which first order differential equations arise. EXAMPLE 1 Mixing At time t = 0 a tank contains Q 0 lb of salt dissolved in 100 gal of water; see Figure…

to the height where it was hit and only slightly longer to reach a point 0.9 m below this height. t = (188 m) /(v0 cos 45°) gives t = 6.21 s, which agrees with this estimate. The ball reaches its maximum height approximately (188 m) / 2 = 94 m from home plate, so at the fence the ball is not far past its maximum height of 47.6 m, so a height of 45.0 m at the fence is reasonable. IDENTIFY: The water moves in projectile motion. SET UP: Let x0 = y0 = 0 and take + y to be positive. ax = 0 ,…

to the height where it was hit and only slightly longer to reach a point 0.9 m below this height. t = (188 m) /(v0 cos 45°) gives t = 6.21 s, which agrees with this estimate. The ball reaches its maximum height approximately (188 m) / 2 = 94 m from home plate, so at the fence the ball is not far past its maximum height of 47.6 m, so a height of 45.0 m at the fence is reasonable. IDENTIFY: The water moves in projectile motion. SET UP: Let x0 = y0 = 0 and take + y to be positive. ax = 0 ,…

to the height where it was hit and only slightly longer to reach a point 0.9 m below this height. t = (188 m) /(v0 cos 45°) gives t = 6.21 s, which agrees with this estimate. The ball reaches its maximum height approximately (188 m) / 2 = 94 m from home plate, so at the fence the ball is not far past its maximum height of 47.6 m, so a height of 45.0 m at the fence is reasonable. IDENTIFY: The water moves in projectile motion. SET UP: Let x0 = y0 = 0 and take + y to be positive. ax = 0 ,…

to the height where it was hit and only slightly longer to reach a point 0.9 m below this height. t = (188 m) /(v0 cos 45°) gives t = 6.21 s, which agrees with this estimate. The ball reaches its maximum height approximately (188 m) / 2 = 94 m from home plate, so at the fence the ball is not far past its maximum height of 47.6 m, so a height of 45.0 m at the fence is reasonable. IDENTIFY: The water moves in projectile motion. SET UP: Let x0 = y0 = 0 and take + y to be positive. ax = 0 ,…

to the height where it was hit and only slightly longer to reach a point 0.9 m below this height. t = (188 m) /(v0 cos 45°) gives t = 6.21 s, which agrees with this estimate. The ball reaches its maximum height approximately (188 m) / 2 = 94 m from home plate, so at the fence the ball is not far past its maximum height of 47.6 m, so a height of 45.0 m at the fence is reasonable. IDENTIFY: The water moves in projectile motion. SET UP: Let x0 = y0 = 0 and take + y to be positive. ax = 0 ,…

to the height where it was hit and only slightly longer to reach a point 0.9 m below this height. t = (188 m) /(v0 cos 45°) gives t = 6.21 s, which agrees with this estimate. The ball reaches its maximum height approximately (188 m) / 2 = 94 m from home plate, so at the fence the ball is not far past its maximum height of 47.6 m, so a height of 45.0 m at the fence is reasonable. IDENTIFY: The water moves in projectile motion. SET UP: Let x0 = y0 = 0 and take + y to be positive. ax = 0 ,…