Recent content by twits
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T
I need a comprehensive answer,
Can anyone describe how the velocities of the particles will change during their movements(with the dragging force in this case) ?- twits
- Post #15
- Forum: Introductory Physics Homework Help
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T
I need a comprehensive answer,
Question: Can we use the known charges, masses of the particles to indicate how far each will go, then just do the problem for 1 particle?- twits
- Post #14
- Forum: Introductory Physics Homework Help
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T
I need a comprehensive answer,
either way is ok. Let's just say it's the opposing force .- twits
- Post #13
- Forum: Introductory Physics Homework Help
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T
I need a comprehensive answer,
Let's just eliminate the medium, will the problem be easier ?- twits
- Post #9
- Forum: Introductory Physics Homework Help
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T
I need a comprehensive answer,
Sum of force on the electron= kQ1*Q2/(r(t))^2 - (v(t))^(1/2) m*v'(t)=kQ1*Q2/(r(t))^2-(v(t))^(1/2) But then how will I connect r(t) and v(t) ? I got stuck here.- twits
- Post #8
- Forum: Introductory Physics Homework Help
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T
Calculating the Velocity of Pin P in a Rod A-Rod B System
Can you give me a picture for illustration ?- twits
- Post #2
- Forum: Introductory Physics Homework Help
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T
I need a comprehensive answer,
Please, anyone? This isn't homework or anything, so if you think you have the answer, just show it ! Thanks.- twits
- Post #6
- Forum: Introductory Physics Homework Help
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T
I need a comprehensive answer,
That's the way I approached the problem, but then I got convoluted with the variables. It's so confusing. Please be more specific !- twits
- Post #4
- Forum: Introductory Physics Homework Help
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T
I need a comprehensive answer,
Anyone ? Please help- twits
- Post #2
- Forum: Introductory Physics Homework Help
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T
I need a comprehensive answer,
I need a comprehensive answer, please help ! I need a comprehensive answer,please help ! I made up this question, but I can not solve it. This bothers me so much. Please try it and explain it to me. Here it is -------------- Conditions: 1 proton, 1 electron, 1 meter away from each...- twits
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- Replies: 19
- Forum: Introductory Physics Homework Help