The magnitude of the vertical velocity vector for an upwardly launched projectile _________. a stays constant b gets smaller and
2 answers:
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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:
1/4 times your earth's weight
Explanation:
assuming the Mass of earth = M
Radius of earth = R
∴ the mass of the planet= 4M
the radius of the planet = 4R
gravitational force of earth is given as =
where G is the gravitational constant
Gravitational force of the planet =
=
=
recall, gravitational force of earth is given as =
∴Gravitational force of planet = 1/4 times the gravitational force of the earth
you would weigh 1/4 times your earth's weight