answer:**Step 1**: Start with the first law of thermodynamics: [ Q = Delta u + A ] Given that: [ Q = 0 ] [ Delta u = c_v (T - T_0) ] [ A = frac{k x^2}{2} ] **Step 2**: Relate the stiffness of the spring to the pressure and area: [ k x = P S ] **Step 3**: Calculate the change in volume (Delta V): [ Delta V = S x ] Since the volume increases by a factor of (n), we use the volume relation: [ V = frac{n}{n-1} S x ] **Step 4**: Use the ideal gas law for one mole of an ideal gas: [ P V = R T ] Substitute (V) into the ideal gas law: [ P frac{n}{n-1} S x = R T ] **Step 5**: Solve for (frac{k x^2}{2}) and express work (A): [ k x^2 = frac{n-1}{n} R T ] [ A = frac{k x^2}{2} = frac{n-1}{n} frac{R T}{2} ] **Step 6**: Substitute (Delta u) and (A) into the first law of thermodynamics: [ c_v (T - T_0) = -frac{n-1}{n} frac{R T}{2} ] Solve for (T): [ c_v T - c_v T_0 = -frac{n-1}{n} frac{R T}{2} ] [ c_v T + frac{n-1}{n} frac{R T}{2} = c_v T_0 ] [ T left( c_v + frac{n-1}{n} frac{R}{2} right) = c_v T_0 ] [ T = frac{T_0}{1 + frac{(n-1) R}{2 n c_v}} ] **Step 7**: Use the ideal gas condition in the initial and final states: [ P_0 V_0 = R T_0 ] [ P cdot n V_0 = R T ] Divide the second equation by the first: [ frac{P cdot n V_0}{P_0 V_0} = frac{R T}{R T_0} ] [ frac{P}{P_0} = frac{T}{n T_0} ] Substitute the expression for (T): [ frac{P}{P_0} = frac{T_0}{n T_0 left(1 + frac{(n-1) R}{2 n c_v}right)} ] [ frac{P}{P_0} = frac{1}{n left(1 + frac{(n-1) R}{2 n c_v}right)} ] **Conclusion**: [ boxed{ P_0 frac{1}{n left(1 + frac{(n-1) R}{2 n c_v} right)} } ]

answer:To find the total number of ways to select 4 runners from 6 for the relay race without any constraints, we use the permutation formula: text{Total ways} = P(6,4) = frac{6!}{(6-4)!} = 6 times 5 times 4 times 3 = 360 Next, we determine the cases where athlete A is running the first leg, we need to select the remaining 3 runners from the remaining 5 athletes: text{Ways A runs first} = P(5,3) = frac{5!}{(5-3)!} = 5 times 4 times 3 = 60 Similarly, for athlete B running the last leg, we need to select the first 3 runners from the remaining 5 athletes: text{Ways B runs last} = P(5,3) = frac{5!}{(5-3)!} = 5 times 4 times 3 = 60 However, there's an overlap where athlete A runs first and athlete B runs last. We need to count this scenario once since it's been counted in both of the above cases. For this overlap, we need to select 2 middle runners from the remaining 4 athletes: text{Overlap ways (A first, B last)} = P(4,2) = frac{4!}{(4-2)!} = 4 times 3 = 12 Now, to find the number of desirable selections where athlete A does not run first and athlete B does not run last, we subtract the selections where A is first, the selections where B is last, and then add back the overlap to avoid double subtraction: text{Desirable selections} = text{Total ways} - text{Ways A runs first} - text{Ways B runs last} + text{Overlap ways (A first, B last)} = 360 - 60 - 60 + 12 = 252 To find the probability that athlete A does not run the first leg and athlete B does not run the last leg, we divide the number of desirable selections by the total number of ways: P = frac{text{Desirable selections}}{text{Total ways}} = frac{252}{360} = frac{7}{10} Thus, the probability is boxed{frac{7}{10}}.

answer:First, we factorize both numbers into their prime factors: [ 105 = 3^1 cdot 5^1 cdot 7^1 ] [ 88 = 2^3 cdot 11^1 ] Next, we identify any common factors between the two numbers. From the prime factorizations, we see: - The prime factors of 105 are 3, 5, and 7. - The prime factors of 88 are 2 and 11. Since there are no common prime factors between 105 and 88, the greatest common divisor is: [ boxed{1} ]

answer:1. **Analyze triangle ABC**: - Since point B is east of A and C is north of B, angle CBA = 90^circ. - Given angle BAC = 30^circ, angle ABC = 90^circ - 30^circ = 60^circ. - triangle ABC is a 30-60-90 triangle. 2. **Determine the sides of triangle ABC**: - The hypotenuse AC = 15sqrt{2}. - In a 30-60-90 triangle, the side opposite the 30-degree angle (AB) is frac{1}{2} the hypotenuse, and the side opposite the 60-degree angle (BC) is frac{sqrt{3}}{2} times the hypotenuse. - Therefore, AB = frac{15sqrt{2}}{2} = 7.5sqrt{2} meters, and BC = frac{15sqrt{2} sqrt{3}}{2} = 7.5sqrt{6} meters. 3. **Consider point D and triangle ADB**: - Point D is 25 meters due north of point C. - Thus, DC = 25 meters. - DB = BC + DC = 7.5sqrt{6} + 25 meters. 4. **Apply the Pythagorean theorem in triangle ADB**: - triangle ADB is a right triangle with AB = 7.5sqrt{2} meters and DB = 7.5sqrt{6} + 25 meters. - AD^2 = AB^2 + DB^2 = (7.5sqrt{2})^2 + (7.5sqrt{6} + 25)^2 = 112.5 + (7.5sqrt{6} + 25)^2. - Simplifying DB^2: (7.5sqrt{6} + 25)^2 = 56.25 cdot 6 + 375sqrt{6} + 625 = 337.5 + 375sqrt{6} + 625. - Hence, AD^2 = 112.5 + 337.5 + 375sqrt{6} + 625 = 1075 + 375sqrt{6}. - Approximating 375sqrt{6} approx 915, AD^2 approx 1990. 5. **Estimate sqrt{1990}**: - We know 44^2 = 1936 and 45^2 = 2025. - Since 1936 < 1990 < 2025, it follows that 44 < sqrt{1990} < 45. **Conclusion with boxed answer**: - The distance AD is between 44 and 45 meters. 44 text{and 45} The final answer is boxed{B) 44 text{and} 45}