Physics, asked by devanshusingh109, 1 month ago

Please Solve this, especially the area of cross section part​

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Answered by nehakalaskar404
0

Answer:

2.5×10-4 ms-1

in a small time interval dt, the number of electrons leaving a cross section is (4×1028)(10−5)vdt.

Therefore, charge crossing in time dt is :

dq=(1.6×10−19)(4×1028)(10−5)vdt

Hence, I=dtdq=6.4×104v=16

This gives: v=2.5×10−4 m/s

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2

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