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:
I think it's the most important part in this
Based on the calculations, the angle through which the tire rotates is equal to 4.26 radians and 244.0 degrees.
<h3>How to calculate the angle?</h3>
In Physics, the distance covered by an object in circular motion can be calculated by using this formula:
S = rθ
<u>Where:</u>
- r is the radius of a circular path.
- θ is the angle measured in radians.
Substituting the given parameters into the formula, we have;
1.87 = 0.44 × θ
θ = 1.87/0.44
θ = 4.26 radians.
Next, we would convert this value in radians to degrees:
θ = 4.26 × 180/π
θ = 4.26 × 180/3.142
θ = 244.0 degrees.
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Answer:
The rate of heat conduction through the layer of still air is 517.4 W
Explanation:
Given:
Thickness of the still air layer (L) = 1 mm
Area of the still air = 1 m
Temperature of the still air ( T) = 20°C
Thermal conductivity of still air (K) at 20°C = 25.87mW/mK
Rate of heat conduction (Q) = ?
To determine the rate of heat conduction through the still air, we apply the formula below.
Q = 517.4 W
Therefore, the rate of heat conduction through the layer of still air is 517.4 W
