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:
496.7 K
Explanation:
The efficiency of a Carnot engine is given by the equation:
where:
is the temperature of the hot reservoir
is the temperature of the cold reservoir
For the engine in the problem, we know that
is the efficiency
is the temperature of the cold reservoir
Solving for , we find:
Answer:
12 mins
Explanation:
The distance covered is 5 km, divide this by 25 to get the fraction of an hour it takes. Doing this you get .2, times this by 60 min (1 hour) to get how many mins it takes
