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Hey , Thanks very much.
Sorry for the wrong answer given by me .
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2 objects with velocity v1 and v2 approach move perpendicularly to each other. Their point of meeting is at a distance l1 and l2 from the objects. What is the shortest distance between the objects ?
Ans : (m1v2 + m2v1)/(v1^2 + v2^2)^1/2
Need explanation.
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Suppose, there are two blocks of distinct masses one over the another. When you put force in the block which is at the top it will experience friction in the opposite direction( very natural ). This frictional force causes the block underneath to move in the direction of the frictional force. Thus the work done by the frictional force on the lower body will be positive. This can be seen when you make a F.B.D.
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Lets consider an example.
Let us take a case of centimeter scale ruler which shows a reading of 1.3 cm.
Here 1 is certain but .3 is doubtful or uncertain. But if the same reading is obtained on a vernier caliper 1.3 would be certain and you would write the reading as 1.30. Here 0 is uncertain . i.e there is no uncertainity. Thus when you write 0 on the right side of decimal digits you actually show the accuracy of your reading.
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let the vectors be A and B. Let R be the resultant vector.
R is equal to A and B.
A^2 + B^2 + 2AB cos = R^2
or A^2 + B^2 + 2AB cos = A^2 ( as R and A are equal )
or B^2 + 2AB cos = 0
or B + 2A cos = 0
or B = -2A cos -----------------------(1)
& A^2 + B^2 + 2AB cos = B^2
or A^2 + 2AB cos = 0
or A + 2B cos = 0
or -4Acos^2 = -A (by (1) )
or cos^2 = 1/4
or cos = 1/2
or = 60 degrees
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