= Joules ÷ (0.5×Kilograms)
14J ÷ 8.5 = 1.64705882
Remember, 1.64705882 = v², so we need to find the square root.
The square root of 1.64705882 is 1.283377894464448
Hope this helps!
To look for displacement, just draw a vector from your beginning stage to your last position and settle for the length of this line. So we begin by drawing a line to the north which is 30 ft, since it is north, the line is going up, then it move 5 ft to the south, so put a line going down, so we are in 25 ft, North so that would be the answer.
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
Explanation:
direction of electric field is same as that of force experienced by the test charge
The wedge and screw simple machines
