Final answer:
The gravitational force exerted by the Moon on an average kilogram of water in a river is calculated using the universal law of gravitation. The force is approximately 0.005 N, which is closest to option A) 0.002 N, although this does not exactly match one of the provided options.
Step-by-step explanation:
To calculate the force exerted by the Moon on an average kilogram of water in a river, we can use the universal law of gravitation, which states that the force between two masses is equal to the gravitational constant times the product of the two masses divided by the square of the distance between the centers of the two masses. The formula is given by:
F = G * (m1 * m2) / r^2
Where:
F is the gravitational force,
G is the gravitational constant (6.67 × 10^-11 N.m^2/kg^2),
m1 is the mass of the Moon (7 × 10^22 kg),
m2 is the mass of the water (1 kg for an average kilogram of water),
r is the distance from the center of the Moon to the water in the river (3 × 10^5 km or 3 × 10^8 m since 1 km = 1000 m).
Plugging in the values we get:
F = (6.67 × 10^-11 N.m^2/kg^2) * (7 × 10^22 kg * 1 kg) / (3 × 10^8 m)^2
F = (6.67 × 10^-11) * (7 × 10^22) / (9 × 10^16)
F ≈ 4.9 × 10^-3 N
The correct force with which the Moon pulls an average kilogram of water in a river is therefore approximately 0.005 N. This value is not one of the options provided, suggesting the possibilities listed may be incorrect or that the question may have an error. However, the option closest to our calculated value is A) 0.002 N.