You are given the equation that is used to solve a problem. For each of these: W

Question

You are given the equation that is used to solve a problem. For each of these: Write a realistic physics problem for which this is the correct equation. Look at worked examples and end-of-chapter problems in the textbook to see what realistic physics problems are like. Finish the solution of the problem. (9.0 times 10^9N middot m^2/C^2)q_1q_2/0.030 m = 9.0 times 10^-5 J q_1 + q_2 = 40 nC (9.0 times 10^9N middot m^2/C^2)(3.0 times 10^-9 C)/r = 18,000 V 1/2(1.67 times 10^-27 kg)v^2_i + 0 = 0 + (9.0 times 10^9 N middot m^2/C^2)(2.0 times 10^-9 C)(1.60 times 10^-19 C)/0.0010 m 400 nC = (100 V)C C = (8.85 times 10^-12 F/m)(0.10 m times 0.10 m)/d

Explanation / Answer

28 ) Equation = Kq1q2/ r ------electrostatic potential energy

( 9 x 10^9) ( q1) ( 40x 10^-9- q1)/ 0.030 = 9 x 10^-5

solving for q1

q1( 40x 10^-9- q1) = 0.03 x 10^ -14

solving this equation will give q1 as apprx equal to 40nC, and q2 very close to 0

29 ) V = kq/r-------electrostatic potential

9 x 10^9 ( 3 x 10^-9) / 18,000 = r

r =18/ 18,000= 0.1 cm

30) 1/2 mv^2 = kq1q2/ r

0.5 ( 1.67 x 10^-27 ) v^ 2= 9 x 10^9 ( 2 x 10^-9) ( 1.60 x 10^-19)/ 0.0010

0.835 x 10^ -8 v^2 = 28800

v = 185.717 x 10^ 4 m/s apprx

31) q= cv

400 x 10^-9 = (8.85 x 10^-12 )( 0.10x 0.10) 100/ d

400 = 8.85 x 10^-3 / d

d= 0.022 mm apprx

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