Answer:
, assuming that the speed of the electron stays the same.
Explanation:
Let denote the speed of this electron. Let denote the electric charge on this electron. Let denote the mass of this electron.
Since the path of this electron is a circle (not a helix,) this path would be in a plane normal to the magnetic field.
Let denote the strength of this magnetic field. The size of the magnetic force on this electron would be:
.
Assuming that there is no other force on this electron. The net force on this electron would be . By Newton's Second Law of motion, the acceleration of this electron would be:
.
On the other hand, since this electron is in a circular motion with a constant speed:
.
Combine the two equations to obtain a relationship between (radius of the path of the electron) and (strength of the magnetic field:)
.
Simplify to obtain:
.
In other words, if the speed of this electron stays the same, the radius of the path of this electron would be inversely proportional to the strength of the magnetic field. Doubling the radius of this path would require halving the strength of the magnetic field (to .)
If the object, ends up with a positive charge, then it is missing electrons. if it is missing electrons, then it must have been removed form the object during the rubbing process.
Answer:
83000 m/s
Explanation:
Using
F = Bqv....................... Equation 1
Where F = Force experienced by the charge, B = magnetic field q = charge of the particle, v = speed of the particle.
make v the subject of the equation
v = F/Bq.................... Equation 2
Given: q = 5.2×10⁻¹⁹ Coulombs, F = 9.5×10⁻¹⁵ Newtons, B = 2.2×10⁻¹ Tesla
Substitute into equation 2
v = 9.5×10⁻¹⁵ /(5.2×10⁻¹⁹×2.2×10⁻¹)
v = 8.3×10⁴ m/s
v = 83000 m/s
Answer:
What evidence supports their claim.
Explanation:
In the end, all that really matters in an experiment, is what scientific evidence supports the claim being made.