bis/view.php?id 4908272 ning. 2/16/2018 11:59 PM010/100 Cakulalor Per odic Table

Question

bis/view.php?id 4908272 ning. 2/16/2018 11:59 PM010/100 Cakulalor Per odic Table Print Question 6 of 10 Sapling Learning A nucleus of the boron-11 isotope consists of five protons and six neutrons. A particular ionized atom of boror-11, whose mass is 1.83 × 10kg, lacks 4 electrons from its neutral state. Find the magnitude and direction of the electric Sield that will levitate this ion, exactly balancing its weight. Take g 9.81 m/s Magnitude: Number N/C Direction Upward O Cannot be O Horizontal determined O Prevous check Answer 0 Next Exit

Explanation / Answer

q= 4e = 1.6 x 4 x10^-19

q = 6.4 x 10^-19 C

applying Fnet = Fe - Fg = 0

q E = m g

(6.4 x 10^-19) (E) = (1.83 x 10^-26) (9.81)

E = 2.805 x 10^-7 N/C . .....Ans

direction -> upward

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