= −1.3×1013C. E25-29 The mass of water is (250 cm3)(1.00 g/cm 3 ) = 250 g. The number of moles of water is (250 g)/(18.0 g/mol) = 13.9 mol. The number of water molecules is (13.9 mol)(6.02×1023mol−1) = 8.37×1024. Resnick Halliday And Krane Solutions![]() Instructors manual for resnick halliday krane ebook it takes me 52 hours just to snag the right download link, and another 9 hours to validate it. Internet could be. Summary of Book: instructor solutions manual for physics by halliday resnick and krane paul stanley beloit college volume 1 chapters 1 24 halliday resnick and. Resnick halliday krane physics volume 1 5th edition physics by halliday resnick and. Instructors solutions manual physics by resnick hallidayinstructor solutions. Click to enlarge Viking 90 Classica Instruction Manual / 13 / /.. Click to enlarge Viking 51 Instruction Manual / 11 / /.. Click to enlarge Viking 49 Special Instruction Manual / 10 / /.. Click to enlarge Viking 78 178 278 Turrisa Instruction Manual / 12 / /.. Viking designer 1 instruction manual. Click to enlarge Viking 33 Instruction Manual / 9 / /.. Each molecule has ten protons, so the total positive charge is Q = (8.37×1024)(10)(1.60×10−19C) = 1.34×107C. E25-30 The total positive charge in 0.250 kg of water is 1.34×107C. Mary’s imbalance is then q1 = (52.0)(4)(1.34×107C)(0.0001) = 2.79×105C, while John’s imbalance is q2 = (90.7)(4)(1.34×107C)(0.0001) = 4.86×105C, The electrostatic force of attraction is then F = 1 4πǫ0 q1q2 r2 = (8.99×109N m2/C2) (2.79×10 5)(4.86×105) (28.0m)2 = 1.6×1018N. 9 E25-31 (a) The gravitational force of attraction between the Moon and the Earth is FG = GMEMM R2, where R is the distance between them. If both the Earth and the moon are provided a charge q, then the electrostatic repulsion would be FE = 1 4πǫ0 q2 R2. Setting these two expression equal to each other, q2 4πǫ0 = GMEMM, which has solution q = √ 4πǫ0GMEMM, = √ 4π(8.85×10−12C2/Nm2)(6.67×10−11Nm2/kg2)(5.98×1024kg)(7.36×1022kg), = 5.71× 1013 C. (b) We need (5.71× 1013 C)/(1.60× 10−19 C) = 3.57× 1032 protons on each body. The mass of protons needed is then (3.57× 1032)(1.67× 10−27 kg) = 5.97× 1065 kg. Ignoring the mass of the electron (why not?) we can assume that hydrogen is all protons, so we need that much hydrogen. P25-1 Assume that the spheres initially have charges q1 and q2. The force of attraction between them is F1 = 1 4πǫ0 q1q2 r212 = −0.108N, where r12 = 0.500m. The net charge is q1 + q2, and after the conducting wire is connected each sphere will get half of the total. The spheres will have the same charge, and repel with a force of F2 = 1 4πǫ0 1 2 (q1 + q2) 1 2 (q1 + q2) r212 = 0.0360N. Since we know the separation of the spheres we can find q1 + q2 quickly, q1 + q2 = 2 √ 4πǫ0r212(0.0360N) = 2.00µC We’ll put this back into the first expression and solve for q2. Resnick Halliday Physics Pdf−0.108N = 1 4πǫ0 (2.00µC− q2)q2 r212, −3.00× 10−12 C2 = (2.00µC− q2)q2, 0 = −q22 + (2.00µC)q2 + (1.73µC)2. The solution is q2 = 3.0µC or q2 = −1.0µC. Then q1 = −1.0µC or q1 = 3.0µC.
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |