Tom's mass is 70 kg and Sally's mass is 50.0 kg. Tom and Sally are standing 20.0 m apart on the dance floor. Sally looks up and sees Tom. She faels an attraction. Supposing that the attraction is gravitational, find its size. 292times 10^11 5.83times 10^-10 B75N 5.83times 10^-11

To find the gravitational attraction between Tom and Sally, we can use Newton's law of universal gravitation:<br /><br />\[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \]<br /><br />where:<br />- \( F \) is the gravitational force,<br />- \( G \) is the gravitational constant, approximately \( 6.674 \times 10^{-11} \, \text{N}\cdot\text{m}^2/\text{kg}^2 \),<br />- \( m_1 \) and \( m_2 \) are the masses of the two objects (Tom and Sally),<br />- \( r \) is the distance between the centers of the two masses.<br /><br />Given:<br />- \( m_1 = 70 \, \text{kg} \) (Tom's mass),<br />- \( m_2 = 50 \, \text{kg} \) (Sally's mass),<br />- \( r = 20 \, \text{m} \).<br /><br />Substitute these values into the formula:<br /><br />\[ F = \frac{6.674 \times 10^{-11} \, \text{N}\cdot\text{m}^2/\text{kg}^2 \times 70 \, \text{kg} \times 50 \, \text{kg}}{(20 \, \text{m})^2} \]<br /><br />\[ F = \frac{6.674 \times 10^{-11} \times 3500}{400} \]<br /><br />\[ F = \frac{2.3359 \times 10^{-7}}{400} \]<br /><br />\[ F = 5.83975 \times 10^{-10} \, \text{N} \]<br /><br />Rounding to three significant figures, the gravitational force is approximately:<br /><br />\[ F \approx 5.84 \times 10^{-10} \, \text{N} \]<br /><br />The closest answer choice is \( 5.83 \times 10^{-10} \).