A. reactants
B. subscript
C. coefficient
D. products
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
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Hello!</h2>
The answer is:
The first option, the walker traveled 360m more than the actual distance between the start and the end points.
Why?
Since each block is 180 m long, we need to calculate the vertical and the horizontal distance, in order to calculate how farther did the travel walk between the start and the end points (displacement).
So, calculating we have:
Traveler:


Actual distance between the start and the end point (displacement):

Now, to calculate how much farter did the traveler walk, we need to use the following equation:

Therefore, we have that distance differnce between the distance covered by the walker and the actual distance is 360m.
Hence, we have that the walker traveled 360m more than the actual distance between the start point and the end point.
Have a nice day!
Thomas Edison is the answer im 100% sure of it.
Explanation:
it is given that, the linear charge density of a charge, 
Firstly, we can define the electric field for a small element and then integrate for the whole. The very small electric field is given by :
..........(1)
The linear charge density is given by :


Integrating equation (1) from x = x₀ to x = infinity



Hence, this is the required solution.