What is Earth's gravitational attraction for a rhinoceros if the rhino has a mass of 750 kg and the Earth's radius (distance) is 6.37times 10^6 meters? The Earth's mass is 5.98times 10^24kg 4.69times 10^10N 7,375 N 7400 N 4.7times 10^10N 4.69times 10^10kg 7,375 kg

Here's how to calculate the Earth's gravitational attraction (force) on the rhinoceros:<br /><br />**1. Recall Newton's Law of Universal Gravitation:**<br /><br />The force of gravity between two objects is given by:<br /><br />F = G * (m1 * m2) / r^2<br /><br />Where:<br /><br />* F is the gravitational force<br />* G is the gravitational constant (6.674 x 10^-11 N⋅m²/kg²)<br />* m1 and m2 are the masses of the two objects<br />* r is the distance between their centers<br /><br />**2. Plug in the values:**<br /><br />* m1 (Earth's mass) = 5.98 x 10^24 kg<br />* m2 (rhino's mass) = 750 kg<br />* r (Earth's radius) = 6.37 x 10^6 m<br />* G = 6.674 x 10^-11 N⋅m²/kg²<br /><br />F = (6.674 x 10^-11 N⋅m²/kg²) * (5.98 x 10^24 kg * 750 kg) / (6.37 x 10^6 m)^2<br /><br />**3. Calculate:**<br /><br />F ≈ 7358.7 N<br /><br />**4. Round to significant figures:**<br /><br />Since the given values have at most three significant figures, we should round our answer accordingly:<br /><br />F ≈ 7360 N<br /><br />**Therefore, the closest answer is 7400 N.** The other options are incorrect, including those with units of kg, as force is measured in Newtons (N).<br />