Answer: 14.52*10^6 m/s
Explanation: In order to explain this problem we have to consider the energy conservation for the electron within the coaxial cylidrical wire.
the change in potential energy for the electron; e*ΔV is equal to energy kinetic gained for the electron so:
e*ΔV=1/2*m*v^2 v^=(2*e*ΔV/m)^1/2= (2*1.6*10^-19*600/9.1*10^-31)^1/2=14.52 *10^6 m/s
Answer:
3.5 N
Explanation:
Let the 0-cm end be the moment point. We know that for the system to be balanced, the total moment about this point must be 0. Let's calculate the moment at each point, in order from 0 to 100cm
- Tension of the string attached at the 0cm end is 0 as moment arm is 0
- 2 N weight suspended from the 10 cm position: 2*10 = 20 Ncm clockwise
- 2 N weight suspended from the 50 cm position: 2*50 = 100 Ncm clockwise
- 1 N stick weight at its center of mass, which is 50 cm position, since the stick is uniform: 1*50 = 50 Ncm clockwise
- 3 N weight suspended from the 60 cm position: 3*60 = 180 Ncm clockwise
- Tension T (N) of the string attached at the 100-cm end: T*100 = 100T Ncm counter-clockwise.
Total Clockwise moment = 20 + 100 + 50 + 180 = 350Ncm
Total counter-clockwise moment = 100T
For this to balance, 100 T = 350
so T = 350 / 100 = 3.5 N
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