An us bomber is flying horizontally at 300 mph at an altitude of 610 m. its target is an iraqi oil tanker crusing 25kph in the s
1 answer:
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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:
Approximately .
Explanation:
The change in the gravitational potential energy of the pendulum is directly related to the change in its height.
Refer to the sketch attached. The pendulum is initially at . Its highest point is at
. The length of segment
gives the change in its height.
The lengths of and
are simply the length of the string,
. To find the length of
, start by calculating the length of
.
forms a leg in the right triangle
. Besides, it is adjacent to the
angle
. Its length would be:
.
The length of would thus be
.
The change in gravitational potential energy can be found with the equation
. In this equation,
is the mass of the object,
near the surface of the earth, and
is the change in the object's height.
In this case, and
. Therefore:
.