A gold wire that is 1.8 mm in diameter and 15 cm long carries a current of 260 mA. How many electrons per second pass a given cr
1 answer:
Answer:
162500000.
Explanation:
Given that
Diameter of the wire , d= 1.8 mm
The length of the wire ,L = 15 cm
Current ,I = 260 m A
The charge on the electron ,e= 1.6 x 10⁻¹⁹ C
We know that Current I is given as
I=Current
q=Charge
t=time
q= I t
q= 260 m t
The total number of electron = n
q= n e
n=162500000 t
The number of electron passe per second will be 162500000.
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Explanation:
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1 )
We shall apply conservation of momentum law to solve the problem.
mv = ( M +m) V , m and M are masses of small and large object , v is the velocity of small object before collision and V is the velocity of both the objects together after collision .
.5 x .2 = (1.5 + .5)V
V = .05 m /s
2 ) We shall use formula for velocity of object after elastic collision as follows
v₁ =
m₁ and m₂ are masses of first and second object u₁ and u₂ are their initial velocity and v₁ and v₂ are their final velocity.
Putting the values
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= - .66 m /s
Since the sign is negative so it will be in opposite direction .