Answer:
Option c) are perpendicular to the electric field
Explanation:
Equipotential surfaces are perpendicular to the electric field. the electric field lines are projected outwards from the equipotential surface, i.e., the lines of the electric field are at 90
to the equipotential surface.
Equipotential surface are those surfaces that have the same potential at any point on the surface. Thus the potential difference at any point on the surface is zero due to same potential.
Any charge particle on this surface will move in a perpendicular direction to the Coulombian force. No work is done by the force on a particle moving on an equipotential surface.
| Impedance | = √ [R² +(ωL)²]
R² = 6800² = 4.624 x 10⁷
(ωL)² = (2 · π · f · 2.3 · 10⁻³)²
= 2.0884 x 10⁻⁴ f²
| Z | = √[ (4.624 x 10⁷) + (2.0884 x 10⁻⁴ f²) ] = 1.6 x 10⁵
(1.6 x 10⁵)² = (4.624 x 10⁷) + (2.0884 x 10⁻⁴ f²)
(2.56 x 10¹⁰) - (4.624 x 10⁷) = 2.0884 x 10⁻⁴ f²
Frequency² = (2.56 x 10¹⁰ - 4.624 x 10⁷) / 2.0884 x 10⁻⁴
= 2.555 x 10¹⁰ / 2.0884 x 10⁻⁴
= 1.224 x 10¹⁴
= 122,400 GHz <== my calculation
11.1 MHz <== online impedance calculator
Obviously, I must have picked up some rounding errors
in the course of my calculation.
Distance, since distance represents how far something has travelled, which would be in our case 2.5m.
Answer:
Explanation:
Gravitational law states that, the force of attraction or repulsion between two masses is directly proportional to the product of the two masses and inversely proportional to the square of their distance apart.
So,
Let the masses be M1 and M2,
F ∝ M1 × M2
Let the distance apart be R
F ∝ 1 / R²
Combining the two equation
F ∝ M1•M2 / R²
G is the constant of proportional and it is called gravitational constant
F = G•M1•M2 / R²
So, to increase the gravitational force, the masses to the object must be increased and the distance apart must be reduced.
So, option c is correct
C. Both objects have large masses and are close together.
