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Please help me

  • Thread starter F.B
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  • #1
F.B
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I am studying for my exam for physics and i came across 2 questions that i sort of need help with.

1. Two identical spheres, with charges of Qa=-2Q and Qb=1Q, attract one another with a force of 9.0 x 10^-4 N. A third identical uncharged sphere C touches B and now C is completely removed from this situation. Finally, the two spheres (A and B) are touched together and move a distance of 2R apart. What is the new force?

I solve this ratio questions using the mx method but thats not my biggest problem. I don't know how to determine the charges of the 2 spheres. Can anyone please help me. I know the charges for both are supposed to be 3/4 for both. But is there a method to determining the charges.

2. A person swims 50 m[E], then 60 m[N40W], then 70 m[S60W], and finally 100 m [W30N]. Determine the resultant displacement.

Some of these angles arent the angle to the horizontal but i get confused, so can you tell me which angles are the angles to the vertical.

Are the ones with angles to the vertical only the 60 m and 70 m because when i draw the 70 m vector the angle seems right, but it cant be right can it?

So please i really need help.
 

Answers and Replies

  • #2
Astronuc
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Science Advisor
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Two identical spheres, with charges of Qa=-2Q and Qb=1Q, attract one another with a force of 9.0 x 10^-4 N.
If one knew the separation associated with the force, then one could determine the charges.

One may assume the spheres have the same radius. Combining one charge sphere with another uncharged sphere of the same size would do what to the charge?


In the vector problem, assume N is vertical (up), W is left and E is right.

You should have some idea of what N40W means in terms of a direction of 40° toward one direction from the other.
 

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