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:
So, the correct answer is <em><u>the strong nuclear force</u></em>. It actually pulls together nuetrons and protons that are in the nucleus. At very tiny distances only, like those inside the nucleus, so, this strong force succeded in dealing with the electromagnetic force, and it basically stops the electrical repulsion of protons from blowing apart the nucleus.
<u><em>Mark as brainlies please, I need a few more :D</em></u>
Answer:
625 W
Explanation:
Applying
P = W/t.................... Equation 1
Where p = power, W = Work, t = time
But,
W = Force (F) × distance (d)
W = Fd........................ Equation 2
Substitute equation 2 into equation 1
P = Fd/t.................... Equation 3
From the question,
Given: F = 5000 N, d = 30 m, t = 4 munites = (4×60) seconds = 240 seconds
Substitute these values into equation 3
P = (5000×30)/240
P = 625 Watt
Answer:
45 s .
Explanation:
The accelerator will first accelerate , then move with uniform velocity and at last it will decelerate to rest .
displacement s = ?
acceleration a = 1 m /s²
Final speed v = 5 m/s
initial speed u = 0
v² = u² + 2as
5² = 0 + 2 x 1 x s
s = 12.5 m
B) Let time of acceleration or deceleration be t
v = u + a t
5 = 0 + 1 t
t = 5 s
Similarly displacement during deceleration = 12.5 m
Total distance during uniform motion = 200 - ( 12.5 + 12.5 ) = 175 m .
velocity of uniform motion = 5 m /s
time during which there was uniform velocity = 175 / 5 = 35 s
Total time = 5 + 35 + 5 = 45 s .
The tangential velocity of the car's tire is the product of the angular velocity and radius of the car's tire which is 11(r) m/s.
<h3>
Angular velocity of the tire</h3>
The angular velocity of the tire is the rate of change of angular displacement of the tire with time.
The magnitude of the angular velocity of the tire is calculated as follows;
ω = 2πN
where;
- N is the number of revolutions per second
ω = 2π x (5.25 / 3)
ω = 11 rad/s
<h3>Tangential velocity of the tire</h3>
The tangential velocity of the car's tire is the product of the angular velocity and radius of the car's tire.
The magnitude of the tangential velocity is caculated as follows;
v = ωr
where;
- r is the radius of the car's tire
v = 11r m/s
Learn more about tangential velocity here: brainly.com/question/25780931
