a) At a position of 2.0m, the Initial energy is
all made up of the potential energy=m*g*hi<span>
and meanwhile at 1.5 all its energy is also potential energy=m*g*hf
The percentage of energy remaining is E=m*g*hi/m*g*hf x 100
and since mass and gravity are constant so it leaves us with
just E=hi/hf
which 1.5/2.0 x100= 75% so we see that we lost 25% of the
energy or 0.25 in fraction
b) Here use the equation vf^2=vi^2+2gd
<span>where g is gravity, vf is the final velocity and vi is the
initial velocity while d is the distance travelled
so in here we are looking for the vi so let us isolate that
variable
we know that at maximum height or peak, the velocity is 0 so
vf is 0
therefore,</span></span>
vi =sqrt(-2gd) <span>
vi =sqrt(-2x-9.81x1.5) </span>
<span>vi =5.4 m/s
<span>c) The energy was converted to heat due to friction with the
air and the ground.</span></span>
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:
the height of the potential energy is 3,200 J
Explanation:
The computation of the kinetic energy is shown below:
Kinetic energy = 1 ÷ 2 × mass × velocity^2
= 1 ÷ 2 × 4 kg × 40 m/s^2
= 3,200 J
Hence the height of the potential energy is 3,200 J
