Answer:
Explanation:
The strength of the gravitational field at the surface of a planet is given by
(1)
where
G is the gravitational constant
M is the mass of the planet
R is the radius of the planet
For the Earth:
For the unknown planet,
Substituting into the eq.(1), we find the gravitational acceleration of planet X relative to that of the Earth:
And substituting g = 9.8 m/s^2,
Answer:
option C
Explanation:
given,
Q = +3.2 x 10⁻¹⁹ C
E = 5.0 X 10⁵ V/m
B = 0.80 T
ion's acceleration is zero
when acceleration is zero the magnitude of both the forces becomes equal.
q E = q V B
v =
v=
v = 6.25 × 10⁵ m/s ≈ 6.3 × 10⁵ m/s
hence, the correct answer is option C
1) Potential difference: 1 V
2)
Explanation:
1)
When a charge moves in an electric field, its electric potential energy is entirely converted into kinetic energy; this change in electric potential energy is given by
where
q is the charge's magnitude
is the potential difference between the initial and final position
In this problem, we have:
is the magnitude of the charge
is the change in kinetic energy of the particle
Therefore, the potential difference (in magnitude) is
2)
Here we have to evaluate the direction of motion of the particle.
We have the following informations:
- The electric potential increases in the +x direction
- The particle is positively charged and moves from point a to b
Since the particle is positively charged, it means that it is moving from higher potential to lower potential (because a positive charge follows the direction of the electric field, so it moves away from the source of the field)
This means that the final position b of the charge is at lower potential than the initial position a; therefore, the potential difference must be negative: