answer:a) To calculate the acceleration of the system, we can use Newton's second law: F = (m_1 + m_2)a where F is the applied force, m1 and m2 are the masses of the crates, and a is the acceleration. Plugging in the given values, we get: 620 N = (60 kg + 130 kg)a a = 1.79 m/s^2 Therefore, the acceleration of the system is 1.79 m/s^2. b) To calculate the force that each crate exerts on the other, we can use the following equation: F_{12} = -F_{21} where F12 is the force exerted by crate 1 on crate 2, and F21 is the force exerted by crate 2 on crate 1. Since the crates are in contact, the force of friction between them is: f = mu_k(m_1 + m_2)g where μk is the coefficient of kinetic friction and g is the acceleration due to gravity. Plugging in the given values, we get: f = 0.15(60 kg + 130 kg)(9.8 m/s^2) f = 283.5 N Therefore, the force that each crate exerts on the other is 283.5 N. c) If the crates are reversed, the total mass of the system will remain the same, so the acceleration will also remain the same. However, the force that each crate exerts on the other will change. To calculate the new force, we can use the following equation: F_{21}' = -F_{12}' where F21' is the force exerted by crate 2 on crate 1, and F12' is the force exerted by crate 1 on crate 2. Plugging in the given values, we get: F_{21}' = -283.5 N F_{12}' = 283.5 N Therefore, the force that crate 2 exerts on crate 1 is 283.5 N, and the force that crate 1 exerts on crate 2 is -283.5 N.

answer:The mean of a set of numbers is found by summing all the numbers and dividing by the count. For the given set, the mean is calculated as follows: Mean = frac{3 sqrt{2} - frac{5}{sqrt{2}}}{2} To simplify, multiply the numerator and denominator by sqrt{2} to get rid of the fraction within the denominator: Mean = frac{3 sqrt{2} cdot sqrt{2} - frac{5}{sqrt{2}} cdot sqrt{2}}{2 cdot sqrt{2}} Mean = frac{6 - 5}{2 cdot sqrt{2}} Mean = frac{1}{2 sqrt{2}} Thus, the mean of the set is frac{1}{2 sqrt{2}}, which can be rationalized by multiplying the numerator and denominator by sqrt{2}: Mean = frac{frac{1}{2 sqrt{2}} cdot sqrt{2}}{sqrt{2}} Mean = frac{frac{sqrt{2}}{2}}{sqrt{2}} Mean = frac{1}{2} So, the mean is frac{1}{2}.

answer:Years | Cashflows | FV factor at 9% | FV of cash flow ---|---|---|---| 1 | 1,210 | 1.411582 | 1,709.14 2 | 1,120 | 1.295029 | 1,450.03 3 | 1,550 | 1.188101 | 1,842.66 4 | 1,910 | 1.090000 | 2,079.90 Total Future Value: 7,081.73 Thus, the future value of the cash flows after 4 years is 7,081.73.

answer:The addition of NaOH will neutralize some of the acid and produce more of the conjugate base. After the addition of NaOH, the number of moles of HF and NaF will be: {eq}n_{HF} = 0.58 mol - 0.14 mol n_{HF} = 0.44 mol {/eq} and {eq}n_{NaF} = 1.23 mol + 0.14 mol n_{NaF} = 1.37 mol {/eq} The concentration of the acid and base will be: {eq}{[HF]} = frac {0.44 mol}{1.07 L} {[HF]} = 0.41 M {/eq} and {eq}{[NaF]} = frac {1.37 mol}{1.07 L} {[NaF]} = 1.28 M {/eq} The pH of the solution will be: {eq}pH = pKa + log(frac {{[NaF]} }{{[HF]} }) pH = 3.456 + log(frac {1.28 M}{0.41 M}) pH = 3.456 + 0.50 pH = 3.96 {/eq} The addition of HCl will neutralize some of the base and produce more of the conjugate acid. After the addition of HCl, the number of moles of HF and NaF will be: {eq}n_{HF} = 0.58 mol + 0.28 mol n_{HF} = 0.86 mol {/eq} and {eq}n_{NaF} = 1.23 mol - 0.28 mol n_{NaF} = 0.95 mol {/eq} The concentration of the acid and base will be: {eq}{[HF]} = frac {0.86 mol}{1.07 L} {[HF]} = 0.80 M {/eq} and {eq}{[NaF]} = frac {0.95 mol}{1.07 L} {[NaF]} = 0.89 M {/eq} The pH of the solution will be: {eq}pH = pKa + log(frac {{[NaF]} }{{[HF]} }) pH = 3.456 + log(frac {0.89 M}{0.80 M}) pH = 3.456 + 0.0045 pH = 3.500 {/eq}