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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A At one constant temp and another at a constant pressure
The outside temperature is one of them, its material another, and the last one is its mass.
The reading of the balance if ,
I ) If the elevator is moving with a steady speed = 50 N
II ) If the elevator is moving upwards with acceleration of 0.2 m / s² = 51 N
T = m g + m a
T = Force
m = Mass
g = Acceleration due to gravity
a = Acceleration
m = 5 kg
g = 10 m / s²
I ) If the elevator is moving with a steady speed,
At steady speed, a = 0
T = ( 5 * 10 ) + ( 5 * 0 )
T = 50 N
II ) If the elevator is moving upwards with acceleration of 0.2 m / s²,
a = 0.2 m / s²
T = ( 5 * 10 ) + ( 5 * 0.2 )
T = 50 + 1
T = 51 N
Therefore, the reading of the balance if ,
I ) If the elevator is moving with a steady speed = 50 N
II ) If the elevator is moving upwards with acceleration of 0.2 m / s² = 51 N
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