Answer:
q = 8.57 10⁻⁵ mC
Explanation:
For this exercise let's use Newton's second law
F = ma
where force is magnetic force
F = q v x B
the bold are vectors, if we write the module of this expression we have
F = qv B sin θ
as the particle moves perpendicular to the field, the angle is θ= 90º
F = q vB
the acceleration of the particle is centripetal
a = v² / r
we substitute
qvB = m v² / r
qBr = m v
q =
The exercise indicates the time it takes in the route that is carried out with constant speed, therefore we can use
v = d / t
the distance is ¼ of the circle,
d =
d =
we substitute
v =
r =
let's calculate
r =
2 2.2 10-3 88 /πpi
r = 123.25 m
let's substitute the values
q =
7.2 10-8 88 / 0.6 123.25
q = 8.57 10⁻⁸ C
Let's reduce to mC
q = 8.57 10⁻⁸ C (10³ mC / 1C)
q = 8.57 10⁻⁵ mC
Answer: A if thats not right its C
Explanation:
Answer: 1.176×10^-3 s
Explanation: The time constant formulae for an RC circuit is given below as
t =RC
Where t = time constant , R = magnitude of resistance = 21 ohms , C = capacitance of capacitor = 56 uf = 56×10^-6 F
t = 56×10^-6 × 21
t = 1176×10^-6
t = 1.176×10^-3 s
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