Ask your own question, for FREE!
Physics 6 Online
OpenStudy (anonymous):

The attraction will vary directly with the separation between the charges. True False <<<< correct?

OpenStudy (anonymous):

"Influence" between bodies is reduced as the bodies are: brought closer moved farther apart <<<< this? increased in charge held stationary

OpenStudy (anonymous):

For the first one, attraction is not directly related with the separation, it is related with square of separation.

OpenStudy (anonymous):

Ok so it is false?

OpenStudy (anonymous):

yep.

OpenStudy (anonymous):

For the second one, you can consider the equation which i write before.. (and also consider it for first question) Distance term between bodies is in the denominator. So if you move farther apart the bodies , the interaction will reduce.

OpenStudy (anonymous):

OK good :) I have another math one again. Given: Q1 = +10 µc = 1.0 x 10-5C Q2 = -15 µc = -1.5 x 10-5 C K = 9.0 x 109 N·m2/C2 r = 6.0 m Determine F. _____ N -2.5 -0.42 -0.23 -0.038 If you can put the numbers in the equation and so I can see how they go in it would help and I could do the math. (easy way copy and paste)

OpenStudy (anonymous):

of course, it was one of the earliest theories einstein came up with its valid for extremely small and extremely large bodies alike (planets ---> molecules)

OpenStudy (anonymous):

\[F=\frac{ (9*10^{9})*(10^{-5})*(-1.5*10^{^{-5}}) }{ 6^{2} } \] you can do this calculation.

OpenStudy (anonymous):

YES I CAN!

OpenStudy (anonymous):

wait what is the number on the 10?

OpenStudy (anonymous):

first 10

OpenStudy (anonymous):

-5

OpenStudy (anonymous):

ohh sorry it must be 9

OpenStudy (anonymous):

the second and third are -5

OpenStudy (anonymous):

Yayyyyy got it its D

OpenStudy (anonymous):

Thank you again!

OpenStudy (anonymous):

you are welcome :)

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!