A charged particle is placed in a space. The particle experiences a force. What can be concluded?
2 answers:
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Answer:
The second vector points due West with a magnitude of 600N
Explanation:
The original vector points with a magnitude of 200N due east, the Resultant vector
points due west (that's how east/west direction can be interpreted, from east to west) with a magnitude of 400N. If we choose East as the positive direction and West as the negative one, we can write the following vectorial equation:
With the negative sign signifying that the vector points west.
(a)
The total mechanical energy of the space probe must be constant, so we can write:
(1)
where
is the kinetic energy at the surface, when the probe is launched
is the gravitational potential energy at the surface
is the final kinetic energy of the probe
is the final gravitational potential energy
Here we have
at the surface, (radius of the planet),
(mass of the planet) and m=10 kg (mass of the probe), so the initial gravitational potential energy is
At the final point, the distance of the probe from the centre of Zero is
so the final potential energy is
So now we can use eq.(1) to find the final kinetic energy:
(b)
The probe reaches a maximum distance of
which means that at that point, the kinetic energy is zero: (the probe speed has become zero):
At that point, the gravitational potential energy is
So now we can use eq.(1) to find the initial kinetic energy: