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Libro Tippens Ejercicios, Ejercicios de Física

Resolución de ejercicios del libro de Tippens

Tipo: Ejercicios

2021/2022

Subido el 18/02/2022

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Chapter 2 Technical Mathematics Physics, 6th Edition
Chapter 2. Technical Mathematics
Signed Numbers
2-1. +7 2-8. -17 2-15. +2 2-22. +12
2-2. +4 2-9. +6 2-16. -2 2-23. +8
2-3. +2 2-10. -32 2-17. -4 2-24. -4
2-4. -2 2-11. -36 2-18. -3 2-25. 0
2-5. -10 2-12. +24 2-19. +2 2-26. +220
2-6. -33 2-13. -48 2-20. -4 2-27. +32
2-7. -5 2-14. +144 2-21. -3 2-28. -32
2-29. (a) -60C; (b) -170C; (c) 360C
2-30. L = 2 mm[(-300C) – (-50C)] = 2 mm(-25) = -50 mm; Decrease in length.
Algebra Review
2-31. x = (2) + (-3) + (-2) = -3; x = -3 2-32. x = (2) – (-3) – (-2) = +7; x = +7
2-33. x = (-3) + (-2) - (+2) = -7; x = -7 2-34. x = -3[(2) – (-2)] = -3(2 + 2) = -12; x = -12
2-35. 2-36.
2-37. x = (-3)2 – (-2)2 = 9 – 4 = 5; x = 5 2-38.
2-39. 2-40. x = (2)2 + (-3)2 + (-2)2; x = 17
2-41. 2-42. x = (2)(-3)[(-2) – (+2)]2; x = -6(-4)2 = -96
2-43. Solve for x: 2ax – b = c; 2ax = b + c;
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Chapter 2. Technical Mathematics

Signed Numbers

2-29. (a) -6^0 C; (b) -17^0 C; (c) 36^0 C 2-30. L = 2 mm[(-30^0 C) – (-5^0 C)] = 2 mm(-25) = -50 mm; Decrease in length. Algebra Review 2-31. x = (2) + (-3) + (-2) = -3; x = -3 2-32. x = (2) – (-3) – (-2) = +7; x = + 2-33. x = (-3) + (-2) - (+2) = -7; x = -7 2-34. x = -3[(2) – (-2)] = -3(2 + 2) = -12; x = - 2-35. 2-. 2-37. x = (-3)^2 – (-2)^2 = 9 – 4 = 5; x = 5 2-. 2-39. 2-40. x = (2)^2 + (-3)^2 + (-2)^2 ; x = 17 2-41. 2-42. x = (2)(-3)[(-2) – (+2)]^2 ; x = -6(-4)^2 = - 2-43. Solve for x: 2ax – b = c; 2ax = b + c;

2-47. 5 m – 16 = 3 m – 4 2-48. 3 p = 7 p - 16 5m – 3 m = -4 + 16 3 p7p = - 2 m = 12; m = 6 -4 p = - 16; p = + 2-49. 4 m = 2( m – 4) 2-50. 3( m – 6) = 6 4 m = 2 m – 8 3m – 18 = 6 2 m = -8; m = -4 3 m = 24; m = + 2-51. 2-52. 2-53. 2-54. 14 = 2( b – 7); 14 = 2b – 14; b = 14 2-55. R^2 = (4)^2 + (3)^2 = 16 + 9 2-56. 3p = 6 + p; p = 3 2-57. V = IR; 2-58. 2-59. 2-60. s = vt + d; d = s – vt 2-61. 2-62. s = ½at^2 ; 2s = at^2 ;

2-108. 0.0000417 2-109. 8.00 x 10^6 2-110. 7.40 x 10^4 2-111. 8.00 x 10^2 2-112. 1.80 x 10-8^ 2-113. 2.68 x 10^9 2-114. 7.40 x 10-3^ 2-115. 1.60x 10-5^ 2-116. 2.70 x 10^19 2-117. 1.80 x 10-3^ 2-118. 2.40 x 10^1 2-119. 2.00 x 10^6 2-120. 2.00 x 10-3^ 2-121. 2.00 x 10-9^ 2-122. 5.71 x 10- 2-123. 2.30 x 10^5 2-124. 6.40x 10^2 2-125. 2.40 x 10^3 2-126. 5.60 x 10-5^ 2-127. –6.90 x 10-2^ 2-128. –3.30 x 10- 2-129. 6.00 x 10-4^ 2-130. 6.40 x 10^6 2-131. –8.00x 10^6 2-132. -4.00 x 10-

Graphs

2-133. Graph of speed vs. time: When t = 4.5 s, v = 144 ft/s; When v = 100 m/s, t = 3.1 s. 2-134. Graph of advance of screw vs. turns: When screw advances 2.75 in., N = 88 turns. 2-135. Graph of wavelength vs. frequency: 350 kHz  857 m; 800 kHz  375 m. 2-136. Electric Power vs. Electric Current: 3.20 A  10.4 W; 8.0 A  64.8 W.

Geometry

2-137. 900. 180^0 , 270^0 , and 45^0 2-138. 2-139a. A = 17^0 , B = 35^0 , C = 38^0 2-139b. A = 50^0 Rule 2; B = 40^0 Rule 2. 2-140a. A = 50^0 Rule 3; B = 130^0 2-140b. B = 70^0 , C = 42^0 Rule 2

A

C

B

D

Right Triangle Trigonometry

2-141. 0.921 2-147. 19.3 2-153. 684 2-159. 54.2^0 2-165. 36.9^0

2-142. 0.669 2-148. 143 2-154. 346 2-160. 6.73^0 2-166. 76.0^0

2-143. 1.66 2-149. 267 2-155. 803 2-161. 50.2^0 2-167. 31.2^0

2-144. 0.559 2-150. 32.4 2-156. 266 2-162. 27.1^0

2-145. 0.875 2-151. 235 2-157. 2191 2-163. 76.8^0

2-146. 0.268 2-152. 2425 2-158. 1620 2-164. 6.37^0

Solve triangles for unknown sides and angles (Exercises 168 – 175): 2-168. tan  = 18/35,  = 35.8^0 ; R = 30.8 ft 2-169. tan  = 600/400,  = 56.3^0 ; R = 721 m. 2-170. y = 650 sin 21^0 = 233 m; x = 650 cos 21^0 = 607 m. 2-171. sin  = 200/500,  = 23.6^0 ; , x = 458 km. 2-172. sin  = 210/400,  = 31.7^0 ; , m = 340 m. 2-173. x = 260 cos 51^0 = 164 in.; y = 260 sin 51^0 = 202 in. 2-174. tan  = 40/80,  = 26.6^0 ; R = 89.4 lb 2-175.  = 180^0 - 120^0 = 60^0 ; y = 300 sin 60^0 = 260 m; x = 300 cos 60^0 = 150 m, left

Challenge Problems

B – A = (8) – (-4) = +12. There is a difference of 24 cm between B – A and A – B. 2-190. Let L = 4Lo; Since , the period will be doubled when the length is quadrupled. Let gm = ge /6, Then, T would be changed by a factor of Thus, the period T on the moon would be 2(2.45) or 4.90 s. 2-191. (a) Area = LW = (3.45 x 10-4^ m)(9.77 x 10-5^ m); Area = 3.37 x 10-8^ m^2. Perimeter (P) = 2L + 2W = 2(L + W); P = 2(3.45 x 10-4^ + 9.77 x 10-5) = 8.85 x 10-4^ m. (b) L = L 0 /2 and W = 2W 0 : A = (L 0 /2)(2W 0 ) = L 0 W 0 ; No change in area. P – P 0 = [2(L 0 /2) + 2(2W 0 )] - [2L 0 + 2W 0 ] = 2W 0 – L 0 P = 2(9.77 x 10-5) – 3.45 x 10-4^ P = -1.50 x 10-4^ m. The area doesn’t change, but the perimeter decreases by 0.150 mm. 2-192. Graph shows when T = 420 K, P = 560 lb/in.^2 ; when T = 600 K, P = 798 lb/in.^2 2-193. Graph shows when V = 26 V, I = 377 mA; when V = 48 V, I = 696 mA.