Answer:
a)
b)
c)
Explanation:
We use the definition of a electric field produced by a point charge:

<u>a)Electric Field due to the alpha particle:</u>

<u>b)Electric Field due to electron:</u>

<u>c)Electric Force on the alpha particle, on the electron:</u>
The alpha particle and electron feel the same force but with opposite direction:

Answer:
A. Zero
Explanation:
The force on a coil of N turns, enclosing an area, A and carrying a current I in the presence of a magnetic field B, is :
F = N * I * A * B * sinθ
Where θ is the angle between the normal of the enclosed area and the magnetic field.
Since the normal of the area is parallel to the magnetic field, θ = 0
Hence:
F = NIABsin0
F = 0 or Zero
Answer:
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Answer:
vp = 0.94 m/s
Explanation
Formula
Vp = position/ time
position: Initial position - Final position
Position = 25 m - (-7 m) = 25 m + 7 m = 32 m
Then
Vp = 32 m / 34 seconds
Vp = 0.94 m/s