The magnitude of the electric field on the master charge is 1.008 x 10¹⁰ N/C, and the force on the test charge is 5.04 x 10⁹ N.
<h3>Electric field on the master charge</h3>
E = kq/r²
where;
- q is magnitude of master charge
- r is distance of separation
- k is Coulomb's constant
E = (9 x 10⁹ x 0.63)/(0.75²)
E = 1.008 x 10¹⁰ N/C
<h3>Force on the test charge</h3>
F = Eq
where;
- E is electric field
- q is the test charge
F = (1.008 x 10¹⁰) x (0.5)
F = 5.04 x 10⁹ N
Thus, the magnitude of the electric field on the master charge is 1.008 x 10¹⁰ N/C, and the force on the test charge is 5.04 x 10⁹ N.
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Answer:
the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh
Explanation:
Given that;
weight of vehicle = 4000 lbs
we know that 1 kg = 2.20462
so
m = 4000 / 2.20462 = 1814.37 kg
Initial velocity
= 60 mph = 26.8224 m/s
Final velocity
= 30 mph = 13.4112 m/s
now we determine change in kinetic energy
Δk =
m(
² -
² )
we substitute
Δk =
×1814.37( (26.8224)² - (13.4112)² )
Δk =
× 1814.37 × 539.5808
Δk = 489500 Joules
we know that; 1 kilowatt hour = 3.6 × 10⁶ Joule
so
Δk = 489500 / 3.6 × 10⁶
Δk = 0.13597 ≈ 0.136 kWh
Therefore, the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh