(2)science who ever gets this right will get a brainlest
2 answers:
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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.
Answer:
power requirement is 23.52 × W
Explanation:
given data
flow rate q = 2 m³/s
elevation h = 1200 m
density of the water ρ = 1000 kg/m³
to find out
power requirement
solution
we will get power by the power equation that is
power = ρ× Q× g× h ...................1
put here all value we get power
power = ρ× Q× g× h
power = 1000 × 2 × 9.8 × 1200
power = 23.52 ×
so power requirement is 23.52 × W