The change in potential energy of the proton is 5.6 x
Joule
<h3>
What is a Uniform Electric Field ?</h3>
A uniform electric field is where the electric field strength is the same at all points in the field. In the uniform field, the force experienced by a charge is the same no matter where it is placed in the field.
Given that a proton moves a distance 10 cm in a uniform electric field of 3.5 kN C, in the direction of the field.
- The distance d = 10 cm = 0.1 m
- Electric field E = 3.5 KN/C
- Proton charge q = 1.6 x
C
The Work done = Fd
but F = Eq
Recall that Electric field E = F/q = V/d
Where V = potential difference.
Let us first calculate the V
E = V/d
V = Ed
Substitute all the parameters into the formula above
V = 3.5 × 10³ × 0.1
V = 350 v
from F/q = V/d
make F the subject of formula and substitute it in work formula
F = Vq/d
W.D = Vq/d x d
W.D = Vq
Substitute all the parameters into the formula above
W.D = 350 x 1.6 x 
W.D = 5.6 x
J
Work done = Energy = Potential Energy
Therefore, the change in potential energy of the proton is 5.6 x
<em> Joule</em>
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Answer:
The speed of the animals is 1.64m/s.
Explanation:
Let us work with variables and call the mass of the two rats
and
, and the length of the rod
.
Using the law of conservation of energy, which says the potential energies of the rats must equal their kinetic energies, we know that when the rod swings to the vertical position,

,
solving for
, we get:

Putting in the values for
,
,
, and
we get:


Therefore, as the rod swings through the vertical position , the speed of the rats is 1.64 m/s.
(a) Determine the circumference of the Earth through the equation,
C = 2πr
Substituting the known values,
C = 2π(1.50 x 10¹¹ m)
C = 9.424 x 10¹¹ m
Then, divide the answer by time which is given to a year which is equal to 31536000 s.
orbital speed = (9.424 x 10¹¹ m)/31536000 s
orbital speed = 29883.307 m/s
Hence, the orbital speed of the Earth is ~29883.307 m/s.
(b) The mass of the sun is ~1.9891 x 10³⁰ kg.
The smallest unit that makes up matter is an atom!
The car's rate of acceleration : a = 2.04 m/s²
<h3>Further explanation</h3>
Given
speed = 110 km/hr
time = 15 s
Required
The acceleration
Solution
110 km/hr⇒30.56 m/s
Acceleration is the change in velocity over time
a = Δv : Δt
Input the value :
a = 30.56 m/s : 15 s
a = 2.04 m/s²