Answer:
B). 3.4 s
Explanation:
As we can see the graph is given between velocity and time
so here we can see that the velocity is changing here with time and initially for some time it moves with constant speed
Then it's speed decreases to next few second and then speed increases to its maximum value
The time after which velocity comes to its maximum value will reach after t = 3 s
so out of the all given options most correct option will be
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