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0
votes
Drawing gravitational equipotential surfaces
You are wrong see potential is inversely proportional to $r$ right
So, $(1/r_1)-(1/r_2)$ is potential difference (there is other constant multiplied to it of course)
Which comes out to be $(r_2)-(r_ …
0
votes
Finding the Tension with Friction and Pulley
Of course tension is same due to constraint ,hence we get 2 equation from free body diagram
MaA = T - f
And
MbA = T + f. (You can take f in any …
0
votes
1
answer
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views
Work done by frictional force [closed]
I used to assume that work done by friction is dependent of path it follows but i am confused as this question or answer of this question suggests that it has nothing to do with give theta
So am i …