The answer is number two, number four, and number one
Answer:
T = 0.0088 m²/s
Explanation:
given,
initial piezometric elevation = 12.5 m
thickness of aquifer = 14 m
discharge = 28.24 L/s = 0.02824 m³/s
we know

k = 0.629 mm/sec
Transmissibilty
T = k × H
T = 0.629 × 14 × 10⁻³
T = 0.0088 m²/s
Answer:
The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C, but the direction is still to the right.
Explanation:
From coulomb's law, F = Eq
Thus,
F = E₁q₁
F = E₂q₂
Then
E₂q₂ = E₁q₁

where;
E₂ is the external electric field due to second test charge = ?
E₁ is the external electric field due to first test charge = 4 x 10⁶ N/C
q₁ is the first test charge = 13 mC
q₂ is the second test charge = 23 mC
Substitute in these values in the equation above and calculate E₂.

The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C when 13 mC test charge is replaced with another test charge of 23 mC.
However, the direction of the external field is still to the right.