static electricity and friction
Answer:
How far will the electron travel beforehitting a plate is 248.125mm
Explanation:
Applying Gauss' law:
Electric Field E = Charge density/epsilon nought
Where charge density=1.0 x 10^-6C/m2 & epsilon nought= 8.85× 10^-12
Therefore E = 1.0 x 10^-6/8.85× 10^-12
E= 1.13×10^5N/C
Force on electron F=qE
Where q=charge of electron=1.6×10^-19C
Therefore F=1.6×10^-19×1.13×10^5
F=1.808×10^-14N
Acceleration on electron a = Force/Mass
Where Mass of electron = 9.10938356 × 10^-31
Therefore a= 1.808×10^-14 /9.11 × 10-31
a= 1.985×10^16m/s^2
Time spent between plate = Distance/Speed
From the question: Distance=1cm=0.01m and speed = 2×10^6m/s^2
Therefore Time = 0.01/2×10^6
Time =5×10^-9s
How far the electron would travel S =ut+ at^2/2 where u=0
S= 1.985×10^16×(5×10^-9)^2/2
S=24.8125×10^-2m
S=248.125mm
Answer:
500 watts
Explanation:
Recall that the definition of power is the amount of energy delivered per unit of time.
In our case, the energy delivered is potential energy which we can estimate as the product of the weight of the object times the distance it is lifted above ground:
200 N x 10 m = 2000 Nm
then the power is the quotient of this potential energy divided the time it took to lift the object to that position:
Power = 2000 / 4 Nm/s = 500 Nm/s = 500 watts
The answer would be solubility
256 kPa because p-guage + p-absolute + p-atmospheric = 256
