Wh Question

c a d a u m; 9. D el eg ue de m a ne ir a qu e o s ubord in a do rece b a ins tr uçõe s so men te de u m a pe ss o a e pres t e contas somente a ela; 10. D efen d a seu s ubor d in ad o q u an do a a u to ri d a de que você atr ibuiu a ele for question ada; 11. D ei xe be m c l aro a s q ue s tõe s sobre a s qu a i s o s su bor din a do s de vem d ec id ir e d e le gue a s d ec is ões a té o meno r ní v el h ier á rqu ic o po s sí ve l; 12. D el eg ue c om con sc iênc i a. N ã o s…

N ⋅ m 2 /C 2 ) (1.06 × 10 −10 C) q = = 42.4 N C 4πε0 r 2 (0.150 m) 2 b) At a distance of 0.300 m from the center (double the sphere’s radius) the field will be 1/4 as strong: 10.6 N C c) Inside the sphere, only the charge inside the radius in question affects the field. In this case, since the radius is half the sphere’s radius, 1/8 of the total charge contributes to the field: (9 × 10 9 N ⋅ m 2 / C 2 ) (1 / 8) (1.06 × 10 −10 C) E= = 21.2 N C (0.075 m) 2 22.23: The point is inside…

N ⋅ m 2 /C 2 ) (1.06 × 10 −10 C) q = = 42.4 N C 4πε0 r 2 (0.150 m) 2 b) At a distance of 0.300 m from the center (double the sphere’s radius) the field will be 1/4 as strong: 10.6 N C c) Inside the sphere, only the charge inside the radius in question affects the field. In this case, since the radius is half the sphere’s radius, 1/8 of the total charge contributes to the field: (9 × 10 9 N ⋅ m 2 / C 2 ) (1 / 8) (1.06 × 10 −10 C) E= = 21.2 N C (0.075 m) 2 22.23: The point is inside…

N ⋅ m 2 /C 2 ) (1.06 × 10 −10 C) q = = 42.4 N C 4πε0 r 2 (0.150 m) 2 b) At a distance of 0.300 m from the center (double the sphere’s radius) the field will be 1/4 as strong: 10.6 N C c) Inside the sphere, only the charge inside the radius in question affects the field. In this case, since the radius is half the sphere’s radius, 1/8 of the total charge contributes to the field: (9 × 10 9 N ⋅ m 2 / C 2 ) (1 / 8) (1.06 × 10 −10 C) E= = 21.2 N C (0.075 m) 2 22.23: The point is inside…

N ⋅ m 2 /C 2 ) (1.06 × 10 −10 C) q = = 42.4 N C 4πε0 r 2 (0.150 m) 2 b) At a distance of 0.300 m from the center (double the sphere’s radius) the field will be 1/4 as strong: 10.6 N C c) Inside the sphere, only the charge inside the radius in question affects the field. In this case, since the radius is half the sphere’s radius, 1/8 of the total charge contributes to the field: (9 × 10 9 N ⋅ m 2 / C 2 ) (1 / 8) (1.06 × 10 −10 C) E= = 21.2 N C (0.075 m) 2 22.23: The point is inside…

N ⋅ m 2 /C 2 ) (1.06 × 10 −10 C) q = = 42.4 N C 4πε0 r 2 (0.150 m) 2 b) At a distance of 0.300 m from the center (double the sphere’s radius) the field will be 1/4 as strong: 10.6 N C c) Inside the sphere, only the charge inside the radius in question affects the field. In this case, since the radius is half the sphere’s radius, 1/8 of the total charge contributes to the field: (9 × 10 9 N ⋅ m 2 / C 2 ) (1 / 8) (1.06 × 10 −10 C) E= = 21.2 N C (0.075 m) 2 22.23: The point is inside…

reduced as g = 9.807 − 3.32 × 10-6 z, with z as the elevation in meters. How many percent is the weight of an airplane reduced when it cruises at 11 000 m? Solution: go= 9.807 ms-2 gH = 9.807 – 3.32 × 10-6 × 11 000 = 9.7705 ms-2 Wo = m g o ; WH = m g H WH/Wo = gH/go = 9.807 = 0.9963 Reduction = 1 – 0.9963 = 0.0037 or 0.37% 9.7705 i.e. we can neglect that for most application Sonntag, Borgnakke and van Wylen 2.31 A car drives at 60 km/h and is brought to a full stop with…

a definite answer to this puzzle. Rather, what we shall do is, first to review the experimental data now available. Then we will move on to describe some of the theoretical Interpretations of the effect, emphasizing those that may bear on the question posed above. Finally we shall draw some conclusions. Because this is essentially a nuclear physics audience I will spend some time in the description and the kinematics of the experiment and its relevant features. This is not intended to…

Therefore x2 = 0, whencex = O. (b) Let xn := 1+ 1/2 + 1/3 + ... + I/n forn EN. Since xn+1 = xn + I/(n + 1) > xn' weseethat (xn) is an increasingsequence.By the . Monotone Convergence Theorem 3.3.2, the question of whether the sequence is convergent or not is reduced to the question of whether the sequence is bounded or not. Attempts to use direct numerical calculations to arrive at a conjecture concerning the possible boundedness of the sequence (xn) lead to inconclusive…

0.4 0.6 0.8 1.0 Input buffer limit Figure 6.10. IBL Throughput vs NI for N T − 40 (approximate) N 9T zero only. Simulation results indicate that for a given topology and traffic pattern there is an optimal input buffer limit size wh ch maximizes throughput for heavy offered load. (2) Another version of IBL flow control scheme was proposed and analyzed by Lam. Each node has a pool of N T buffers, all of which may be occupied by transit traffic. However, no more than N I (…