Answer:
Option B. Decreases
Explanation:
Coulomb's law states that:
F = Kq₁q₂ / r²
Where:
F => is the force of attraction between two charges
K => is the electrical constant.
q₁ and q₂ => are the two charges
r => is the distance apart.
From the formula:
F = Kq₁q₂ / r²
The force of attraction (F) is inversely proportional to the square of their separating distance (r).
This implies that as the distance between them increase, the force of attraction between the two charges will decrease and as the distance between two charges decrease, the force of attraction between them will increase.
Considering the question given above and the illustration given above, the force of attraction will decrease as their distance of separation increases.
Option B gives the right answer to the question.
Explanation:
If force or distance is increased, then amount of workdone will also increase.
Answer:
Magnitude of the force is

direction of the force is given as
West of South
Explanation:
As we know that force is a vector quantity and in order to find the resultant of two or more forces we need to add them vectorialy
So here we have

here we know that first force is of magnitude 2 N towards east

second force is also of 2.0 N due North

now from above equation


so magnitude of the force is given as


direction of the force is given as


West of South
Answer:
The maximum wavelength of light that could liberate electrons from the aluminum metal is 303.7 nm
Explanation:
Given;
wavelength of the UV light, λ = 248 nm = 248 x 10⁻⁹ m
maximum kinetic energy of the ejected electron, K.E = 0.92 eV
let the work function of the aluminum metal = Ф
Apply photoelectric equation:
E = K.E + Ф
Where;
Ф is the minimum energy needed to eject electron the aluminum metal
E is the energy of the incident light
The energy of the incident light is calculated as follows;

The work function of the aluminum metal is calculated as;
Ф = E - K.E
Ф = 8.02 x 10⁻¹⁹ - (0.92 x 1.602 x 10⁻¹⁹)
Ф = 8.02 x 10⁻¹⁹ J - 1.474 x 10⁻¹⁹ J
Ф = 6.546 x 10⁻¹⁹ J
The maximum wavelength of light that could liberate electrons from the aluminum metal is calculated as;