A gardener pushes her 90 kg wheelbarrow in a straight line out of her garage and across the yard in 18 s. If her average velocit
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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:
t = 5.59x10⁴ y
Explanation:
To calculate the time for the ¹⁴C drops to 1.02 decays/h, we need to use the next equation:
(1)
<em>where : is the number of decays with time, A₀: is the initial activity, λ: is the decay constant and t: is the time.</em>
To find A₀ we can use the following equation:
(2)
<em>where N₀: is the initial number of particles of ¹⁴C in the 1.03g of the trees carbon </em>
From equation (2), the N₀ of the ¹⁴C in the trees carbon can be calculated as follows:
<em>where : is the tree's carbon mass,
: is the Avogadro's number and
: is the ¹²C mass. </em>
Similarly, from equation (2) λ is:
<em>where t 1/2: is the half-life of ¹⁴C= 5700 years </em>
So, the initial activity A₀ is:
Finally, we can calculate the time from equation (1):
I hope it helps you!
Answer:
The following vectors have been given:
The angle between these two vectors can be found by: