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物理計算題 (質量)

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Question 1On their wedding day, my father gave my mother a gold ring. It weighted 3.70g.On their golden wedding anniversary, 50 years later, it weighted 3.30g.How any atoms were lost each second, on average, from the ring during that time?The mass of a gold atom is 3.29 X 10^ -27kg.There are 86400... 顯示更多 Question 1 On their wedding day, my father gave my mother a gold ring. It weighted 3.70g. On their golden wedding anniversary, 50 years later, it weighted 3.30g. How any atoms were lost each second, on average, from the ring during that time? The mass of a gold atom is 3.29 X 10^ -27kg. There are 86400 second in a day and 365.25 days in a year Question 2 According to Newton's law of universal gravitation is represented by F=G/(Mm/r^2) Here F is the gravitational interaction force acting between any two objects of mass M and m whose separation, centre to centre, is a distance r. (a) The SI unit of force is the 'newton', N. in terms of fundamental units, it is kgms^-2, kilogram metre per sencond squared. (b) The gravitational force between a pair of 1 kg masses 1 m apart is measured as 6.7 X 10^ -11 N. What is the value of the proportionality constant G? (c) My mass is 90 kg and I'm usually 6380 km from Earth's mass is 6.0 X 10^24 kg. Calculate my weight, the gravitational force exerted by Earth on me. 請列算式及步驟, thank you

最佳解答:

1. Let n be the number of atoms lost in 1 second. Hence, no. of atoms lost in 50 years = (50 x 365.25 x 86400)n Mass of atoms lost in 50 years = (50 x 365.25 x 86400)n x 3.29x10^-27 kg Since the ring has lost a mass of (3.7 - 3.3) g = 0.4 g = 4x10^-4 kg in 50 years thus, (50 x 365.25 x 86400)n x 3.29x10^-27 = 4x10^-4 i.e. n = 7.71 x 10^13 atoms 2. (a) what is the question ? (b) Using F = GMm/r^2 6.7x10^-11 = G(1 x 1)/1^2 G = 6.7 x 10^-11 N.m^2/kg^2 (c) Weight = ((6.7x10^-11) x 90 x 6x10^24)/(6380x10^3)^2 N = 888.8 N

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更新2a問題 What arethe SI units of the proporDonality constant G?5AEEC82B53E1B405