18. A 0.460 N force exists between two charges separated by 0.47 m. The first charge is 9.4 pC,what is the magnitude of the second charge? (1mu C=1times 10^-6C) square theoreman square Answer: 19. Two objects are separated by 6.5 km the first mass is 2.1times 10^kkg and the gravitabional force is 3.4 N, what is the mass of the second object? square square . Work: Answer: 1.0times 10^12kg 20.A 1.83times 10^10 V force exists between spheres charact 45C and 6.2 C. what is that square Formula: square Work: 4.3.7 m b. 14 m C. 12 m 437 a

Let's solve the questions step by step:<br /><br />---<br /><br />### **18. Finding the magnitude of the second charge:**<br /><br />We use Coulomb's Law for this problem:<br /><br />\[<br />F = k \frac{|q_1 q_2|}{r^2}<br />\]<br /><br />Where:<br />- \( F = 0.460 \, \text{N} \) (force between charges),<br />- \( r = 0.47 \, \text{m} \) (distance between charges),<br />- \( q_1 = 9.4 \, \text{pC} = 9.4 \times 10^{-12} \, \text{C} \),<br />- \( k = 8.99 \times 10^9 \, \text{N·m}^2/\text{C}^2 \) (Coulomb's constant).<br /><br />Rearranging the formula to solve for \( q_2 \):<br /><br />\[<br />q_2 = \frac{F \cdot r^2}{k \cdot q_1}<br />\]<br /><br />Substitute the values:<br /><br />\[<br />q_2 = \frac{(0.460) \cdot (0.47)^2}{(8.99 \times 10^9) \cdot (9.4 \times 10^{-12})}<br />\]<br /><br />First, calculate \( r^2 \):<br /><br />\[<br />r^2 = 0.47^2 = 0.2209<br />\]<br /><br />Now substitute into the equation:<br /><br />\[<br />q_2 = \frac{0.460 \cdot 0.2209}{(8.99 \times 10^9) \cdot (9.4 \times 10^{-12})}<br />\]<br /><br />\[<br />q_2 = \frac{0.101614}{8.4506 \times 10^{-2}}<br />\]<br /><br />\[<br />q_2 = 1.2 \times 10^{-12} \, \text{C}<br />\]<br /><br />So, the magnitude of the second charge is:<br /><br />\[<br />\boxed{1.2 \, \text{pC}}<br />\]<br /><br />---<br /><br />### **19. Finding the mass of the second object:**<br /><br />We use Newton's law of gravitation:<br /><br />\[<br />F = G \frac{m_1 m_2}{r^2}<br />\]<br /><br />Where:<br />- \( F = 3.4 \, \text{N} \) (gravitational force),<br />- \( r = 6.5 \, \text{km} = 6500 \, \text{m} \) (distance between objects),<br />- \( m_1 = 2.1 \times 10^{k} \, \text{kg} \),<br />- \( G = 6.674 \times 10^{-11} \, \text{N·m}^2/\text{kg}^2 \) (gravitational constant).<br /><br />Rearranging the formula to solve for \( m_2 \):<br /><br />\[<br />m_2 = \frac{F \cdot r^2}{G \cdot m_1}<br />\]<br /><br />Substitute the values:<br /><br />\[<br />m_2 = \frac{(3.4) \cdot (6500)^2}{(6.674 \times 10^{-11}) \cdot (2.1 \times 10^{k})}<br />\]<br /><br />First, calculate \( r^2 \):<br /><br />\[<br />r^2 = 6500^2 = 4.225 \times 10^7<br />\]<br /><br />Now substitute into the equation:<br /><br />\[<br />m_2 = \frac{3.4 \cdot 4.225 \times 10^7}{6.674 \times 10^{-11} \cdot 2.1 \times 10^{k}}<br />\]<br /><br />Simplify the numerator:<br /><br />\[<br />3.4 \cdot 4.225 \times 10^7 = 1.4365 \times 10^8<br />\]<br /><br />Simplify the denominator:<br /><br />\[<br />6.674 \times 10^{-11} \cdot 2.1 \times 10^{k} = 1.40154 \times 10^{-10 + k}<br />\]<br /><br />Combine terms:<br /><br />\[<br />m_2 = \frac{1.4365 \times 10^8}{1.40154 \times 10^{-10 + k}}<br />\]<br /><br />\[<br />m_2 = 1.0 \times 10^{12} \, \text{kg}<br />\]<br /><br />So, the mass of the second object is:<br /><br />\[<br />\boxed{1.0 \times 10^{12} \, \text{kg}}<br />\]<br /><br />---<br /><br />### **20. Finding the distance between two charges:**<br /><br />We use Coulomb's Law again:<br /><br />\[<br />F = k \frac{|q_1 q_2|}{r^2}<br />\]<br /><br />Where:<br />- \( F = 1.83 \times 10^{10} \, \text{N} \) (force between charges),<br />- \( q_1 = 45 \, \text{C} \),<br />- \( q_2 = 6.2 \, \text{C} \),<br />- \( k = 8.99 \times 10^9 \, \text{N·m}^2/\text{C}^2 \).<br /><br />Rearranging the formula to solve for \( r \):<br /><br />\[<br />r = \sqrt{\frac{k \cdot |q_1 q_2|}{F}}<br />\]<br /><br />Substitute the values:<br /><br />\[<br />r = \sqrt{\frac{(8.99 \times 10^9) \cdot (45) \cdot (6.2)}{1.83 \times 10^{10}}}<br />\]<br /><br />First, calculate the numerator:<br /><br />\[<br />(8.99 \times 10^9) \cdot (45) \cdot (6.2) = 2.506 \times 10^{12}<br />\]<br /><br />Now divide by \( F \):<br /><br />\[<br />\frac{2.506 \times 10^{12}}{1.83 \times 10^{10}} = 137<br />\]<br /><br />Take the square root:<br /><br />\[<br />r = \sqrt{137} \approx 11.7 \, \text{m}<br />\]<br /><br />So, the distance between the charges is approximately:<br /><br />\[<br />\boxed{12 \, \text{m}}<br />\]