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:
The value is
Explanation:
From the we are told that
The initial speed of the object is
The greatest height it reached is 
Generally from kinematic equation we have that

At maximum height v = 0 m/s
So

=> 
Here H is the height from the initial height to the maximum height
So the initial height is mathematically represented as

=> 
=> 
Generally the time taken for the object to reach maximum height is mathematically evaluated using kinematic equation as follows

At maximum height v = 0 m/s

=> 
Generally the time taken for the object to move from the maximum height to the ground is mathematically using kinematic equation as follows

Here the initial velocity is 0 m/s given that its the velocity at maximum height
Also g is positive because we are moving in the direction of gravity
So

=> 
Generally the total time taken is mathematically represented as

=> 
=>
The gravitational force is gravity meanwhile the electric force is electric