今天发现，ChatGPT免费版可以正确解答这道AP Physics 1流体力学文字题3小问里面的第1问也就是Part a了。它依然无法理解Part b和c，但是哪天它理解了Part b和c，就意味着它能够在完全没有图片信息的情况下完整构建真实场景，我这个tutor100%将失业。这一天也许3年才到来，也许半年就到来。 You fill a 20-cm-tall container with glycerol so that the glycerol (density 1260 kg/m3) reaches the 10-cm mark. You place an oak cube (density 900 kg/m3) into the container. Each side of the cube is 10 cm. (a) What is the distance x from the upper face of the cube to the glycerol surface assuming the cube is in an upright position? (b) Make a prediction using qualitative reasoning (without mathematics) about what will happen to the cube in (a) if you now add olive oil (density 800 kg/m3) to the container until it is completely full. Will the distance x decrease, increase, or stay the same? (c) Determine the new distance x using mathematics (physics laws) and compare your result with the prediction in (b). 最新更新：今天网友们告知Copilot和ChatGPT o1收费版已经能够正确解答这题所有步骤！大家可能不知道这题(b)(c)两部分有多么困难。昨天我辅导TJ的2名AP Physics 1学生基本流体力学，她们Digital SAT数学都是790或800分，在我引入阿基米德原理以后，她们都能正确解出(a), 但她们都画不出(b)(c)的示意图，完全搞不懂受力分析。当我告诉她们怎么画图怎么做受力分析怎么解方程以后，她们花了30分钟才弄明白。而Copilot和ChatGPT o1已经打败几乎所有Digital SAT数学接近满分学生。而我也已经接近失业。
from Copilot: Final Answer for (a) The distance ( x ) from the upper face of the cube to the glycerol surface is 2.85 cm. (b) Here''s what would happen qualitatively: When you add olive oil (density ( 800 , \text{kg/m}^3 )) to the container, it will form a layer above the glycerol (density ( 1260 , \text{kg/m}^3 )) because olive oil is less dense and cannot mix with glycerol easily. The oak cube, which floats partly submerged due to glycerol''s buoyant force, will now interact with both the glycerol and olive oil. The olive oil provides a weaker buoyant force compared to glycerol, as its density is less than both glycerol and the oak cube. This means the glycerol will continue to bear most of the load of the cube, but the addition of the olive oil on top will exert an additional, albeit smaller, buoyant force. As a result, the cube will float higher overall, causing the distance ( x ) (from the cube''s upper face to the glycerol surface) to increase. The olive oil essentially reduces how much of the cube remains submerged in the glycerol, lifting the cube slightly farther above the glycerol''s surface. In short: The distance ( x ) will increase. Let me know if you''d like further clarification! (c) Determine the new distance x using mathematics (physics laws) and compare your result with the prediction in (b). To calculate the new distance ( x ), we consider the combined buoyant forces from glycerol and olive oil acting on the oak cube. Here''s a step-by-step breakdown: 1. Buoyant Force Analysis The cube now displaces glycerol and olive oil. The submerged height in glycerol is denoted ( h_{\text{glycerol}} ), and the submerged height in olive oil is ( h_{\text{oil}} ). Together: [ h_{\text{glycerol}} + h_{\text{oil}} = 0.1 , \text{m} , ( \text{total height of the cube} ) ] 2. Equilibrium Condition At equilibrium, the buoyant force equals the weight of the cube. The buoyant force is the sum of the forces due to glycerol and olive oil: [ F_{\text{buoyant}} = F_{\text{glycerol}} + F_{\text{oil}} ] The weight of the cube is: [ F_{\text{cube}} = \rho_{\text{oak}} \cdot V_{\text{cube}} \cdot g = 8.82 , \text{N} , (\text{as calculated earlier}). ] For glycerol: [ F_{\text{glycerol}} = \rho_{\text{glycerol}} \cdot V_{\text{glycerol displaced}} \cdot g ] The displaced glycerol volume is: [ V_{\text{glycerol displaced}} = \text{base area of cube} \cdot h_{\text{glycerol}} = 0.01 \cdot h_{\text{glycerol}} , \text{m}^3 ] So: [ F_{\text{glycerol}} = 1260 \cdot (0.01 \cdot h_{\text{glycerol}}) \cdot 9.8 = 123.48 \cdot h_{\text{glycerol}} , \text{N} ] For olive oil: [ F_{\text{oil}} = \rho_{\text{oil}} \cdot V_{\text{oil displaced}} \cdot g = 800 \cdot (0.01 \cdot h_{\text{oil}}) \cdot 9.8 = 78.4 \cdot h_{\text{oil}} , \text{N} ] 3. Solve for ( h_{\text{glycerol}} ) and ( h_{\text{oil}} ) Combine the forces: [ F_{\text{cube}} = F_{\text{glycerol}} + F_{\text{oil}} ] Substitute: [ 8.82 = 123.48 \cdot h_{\text{glycerol}} + 78.4 \cdot h_{\text{oil}} ] Since ( h_{\text{glycerol}} + h_{\text{oil}} = 0.1 ), we substitute ( h_{\text{oil}} = 0.1 - h_{\text{glycerol}} ): [ 8.82 = 123.48 \cdot h_{\text{glycerol}} + 78.4 \cdot (0.1 - h_{\text{glycerol}}) ] Simplify: [ 8.82 = 123.48 \cdot h_{\text{glycerol}} + 7.84 - 78.4 \cdot h_{\text{glycerol}} ] [ 8.82 - 7.84 = (123.48 - 78.4) \cdot h_{\text{glycerol}} ] [ 0.98 = 45.08 \cdot h_{\text{glycerol}} ] Solve for ( h_{\text{glycerol}} ): [ h_{\text{glycerol}} = \frac{0.98}{45.08} \approx 0.0217 , \text{m} = 2.17 , \text{cm}. ] Now, calculate ( h_{\text{oil}} ): [ h_{\text{oil}} = 0.1 - h_{\text{glycerol}} = 0.1 - 0.0217 = 0.0783 , \text{m} = 7.83 , \text{cm}. ] 4. Determine the Distance ( x ) The portion of the cube above the glycerol surface is: [ x = h_{\text{oil}} = 7.83 , \text{cm}. ] Final Answer The new distance ( x ) is 7.83 cm. Compared to the earlier result (( x = 2.85 , \text{cm} )), this confirms that ( x ) increases, aligning with the qualitative prediction in part (b). The addition of olive oil raises the cube further out of the glycerol due to its lower density.
from Copilot: Final Answer for (a) The distance ( x ) from the upper face of the cube to the glycerol surface is 2.85 cm. (b) Here''s what would happen qualitatively: When you add olive oil (density ( 800 , \text{kg/m}^3 )) to the container, it will form a layer above the glycerol (density ( 1260 , \text{kg/m}^3 )) because olive oil is less dense and cannot mix with glycerol easily. The oak cube, which floats partly submerged due to glycerol''s buoyant force, will now interact with both the glycerol and olive oil. The olive oil provides a weaker buoyant force compared to glycerol, as its density is less than both glycerol and the oak cube. This means the glycerol will continue to bear most of the load of the cube, but the addition of the olive oil on top will exert an additional, albeit smaller, buoyant force. As a result, the cube will float higher overall, causing the distance ( x ) (from the cube''s upper face to the glycerol surface) to increase. The olive oil essentially reduces how much of the cube remains submerged in the glycerol, lifting the cube slightly farther above the glycerol''s surface. In short: The distance ( x ) will increase. Let me know if you''d like further clarification! (c) Determine the new distance x using mathematics (physics laws) and compare your result with the prediction in (b). To calculate the new distance ( x ), we consider the combined buoyant forces from glycerol and olive oil acting on the oak cube. Here''s a step-by-step breakdown: 1. Buoyant Force Analysis The cube now displaces glycerol and olive oil. The submerged height in glycerol is denoted ( h_{\text{glycerol}} ), and the submerged height in olive oil is ( h_{\text{oil}} ). Together: [ h_{\text{glycerol}} + h_{\text{oil}} = 0.1 , \text{m} , ( \text{total height of the cube} ) ] 2. Equilibrium Condition At equilibrium, the buoyant force equals the weight of the cube. The buoyant force is the sum of the forces due to glycerol and olive oil: [ F_{\text{buoyant}} = F_{\text{glycerol}} + F_{\text{oil}} ] The weight of the cube is: [ F_{\text{cube}} = \rho_{\text{oak}} \cdot V_{\text{cube}} \cdot g = 8.82 , \text{N} , (\text{as calculated earlier}). ] For glycerol: [ F_{\text{glycerol}} = \rho_{\text{glycerol}} \cdot V_{\text{glycerol displaced}} \cdot g ] The displaced glycerol volume is: [ V_{\text{glycerol displaced}} = \text{base area of cube} \cdot h_{\text{glycerol}} = 0.01 \cdot h_{\text{glycerol}} , \text{m}^3 ] So: [ F_{\text{glycerol}} = 1260 \cdot (0.01 \cdot h_{\text{glycerol}}) \cdot 9.8 = 123.48 \cdot h_{\text{glycerol}} , \text{N} ] For olive oil: [ F_{\text{oil}} = \rho_{\text{oil}} \cdot V_{\text{oil displaced}} \cdot g = 800 \cdot (0.01 \cdot h_{\text{oil}}) \cdot 9.8 = 78.4 \cdot h_{\text{oil}} , \text{N} ] 3. Solve for ( h_{\text{glycerol}} ) and ( h_{\text{oil}} ) Combine the forces: [ F_{\text{cube}} = F_{\text{glycerol}} + F_{\text{oil}} ] Substitute: [ 8.82 = 123.48 \cdot h_{\text{glycerol}} + 78.4 \cdot h_{\text{oil}} ] Since ( h_{\text{glycerol}} + h_{\text{oil}} = 0.1 ), we substitute ( h_{\text{oil}} = 0.1 - h_{\text{glycerol}} ): [ 8.82 = 123.48 \cdot h_{\text{glycerol}} + 78.4 \cdot (0.1 - h_{\text{glycerol}}) ] Simplify: [ 8.82 = 123.48 \cdot h_{\text{glycerol}} + 7.84 - 78.4 \cdot h_{\text{glycerol}} ] [ 8.82 - 7.84 = (123.48 - 78.4) \cdot h_{\text{glycerol}} ] [ 0.98 = 45.08 \cdot h_{\text{glycerol}} ] Solve for ( h_{\text{glycerol}} ): [ h_{\text{glycerol}} = \frac{0.98}{45.08} \approx 0.0217 , \text{m} = 2.17 , \text{cm}. ] Now, calculate ( h_{\text{oil}} ): [ h_{\text{oil}} = 0.1 - h_{\text{glycerol}} = 0.1 - 0.0217 = 0.0783 , \text{m} = 7.83 , \text{cm}. ] 4. Determine the Distance ( x ) The portion of the cube above the glycerol surface is: [ x = h_{\text{oil}} = 7.83 , \text{cm}. ] Final Answer The new distance ( x ) is 7.83 cm. Compared to the earlier result (( x = 2.85 , \text{cm} )), this confirms that ( x ) increases, aligning with the qualitative prediction in part (b). The addition of olive oil raises the cube further out of the glycerol due to its lower density. baidukaohe333 发表于 2025-03-23 02:51
xiaxie8 发表于 2025-03-23 01:17 今天发现，ChatGPT免费版可以正确解答这道AP Physics 1流体力学文字题3小问里面的第1问也就是Part a了。它依然无法理解Part b和c，但是哪天它理解了Part b和c，就意味着它能够在完全没有图片信息的情况下完整构建真实场景，我这个tutor100%将失业。这一天也许3年才到来，也许半年就到来。 You fill a 20-cm-tall container with glycerol so that the glycerol (density 1260 kg/m3) reaches the 10-cm mark. You place an oak cube (density 900 kg/m3) into the container. Each side of the cube is 10 cm. (a) What is the distance x from the upper face of the cube to the glycerol surface assuming the cube is in an upright position? (b) Make a prediction using qualitative reasoning (without mathematics) about what will happen to the cube in (a) if you now add olive oil (density 800 kg/m3) to the container until it is completely full. Will the distance x decrease, increase, or stay the same? (c) Determine the new distance x using mathematics (physics laws) and compare your result with the prediction in (b). 最新更新：今天网友们告知Copilot和ChatGPT o1收费版已经能够正确解答这题所有步骤！大家可能不知道这题(b)(c)两部分有多么困难。昨天我辅导TJ的2名AP Physics 1学生基本流体力学，她们Digital SAT数学都是790或800分，在我引入阿基米德原理以后，她们都能正确解出(a), 但她们都画不出(b)(c)的示意图，完全搞不懂受力分析。当我告诉她们怎么画图怎么做受力分析怎么解方程以后，她们花了30分钟才弄明白。而Copilot和ChatGPT o1已经打败几乎所有Digital SAT数学接近满分学生。而我也已经接近失业。
baidukaohe333 发表于 2025-03-23 02:51 from Copilot: Final Answer for (a) The distance ( x ) from the upper face of the cube to the glycerol surface is 2.85 cm. (b) Here''s what would happen qualitatively: When you add olive oil (density ( 800 , \text{kg/m}^3 )) to the container, it will form a layer above the glycerol (density ( 1260 , \text{kg/m}^3 )) because olive oil is less dense and cannot mix with glycerol easily. The oak cube, which floats partly submerged due to glycerol''s buoyant force, will now interact with both the glycerol and olive oil. The olive oil provides a weaker buoyant force compared to glycerol, as its density is less than both glycerol and the oak cube. This means the glycerol will continue to bear most of the load of the cube, but the addition of the olive oil on top will exert an additional, albeit smaller, buoyant force. As a result, the cube will float higher overall, causing the distance ( x ) (from the cube''s upper face to the glycerol surface) to increase. The olive oil essentially reduces how much of the cube remains submerged in the glycerol, lifting the cube slightly farther above the glycerol''s surface. In short: The distance ( x ) will increase. Let me know if you''d like further clarification! (c) Determine the new distance x using mathematics (physics laws) and compare your result with the prediction in (b). To calculate the new distance ( x ), we consider the combined buoyant forces from glycerol and olive oil acting on the oak cube. Here''s a step-by-step breakdown: 1. Buoyant Force Analysis The cube now displaces glycerol and olive oil. The submerged height in glycerol is denoted ( h_{\text{glycerol}} ), and the submerged height in olive oil is ( h_{\text{oil}} ). Together: [ h_{\text{glycerol}} + h_{\text{oil}} = 0.1 , \text{m} , ( \text{total height of the cube} ) ] 2. Equilibrium Condition At equilibrium, the buoyant force equals the weight of the cube. The buoyant force is the sum of the forces due to glycerol and olive oil: [ F_{\text{buoyant}} = F_{\text{glycerol}} + F_{\text{oil}} ] The weight of the cube is: [ F_{\text{cube}} = \rho_{\text{oak}} \cdot V_{\text{cube}} \cdot g = 8.82 , \text{N} , (\text{as calculated earlier}). ] For glycerol: [ F_{\text{glycerol}} = \rho_{\text{glycerol}} \cdot V_{\text{glycerol displaced}} \cdot g ] The displaced glycerol volume is: [ V_{\text{glycerol displaced}} = \text{base area of cube} \cdot h_{\text{glycerol}} = 0.01 \cdot h_{\text{glycerol}} , \text{m}^3 ] So: [ F_{\text{glycerol}} = 1260 \cdot (0.01 \cdot h_{\text{glycerol}}) \cdot 9.8 = 123.48 \cdot h_{\text{glycerol}} , \text{N} ] For olive oil: [ F_{\text{oil}} = \rho_{\text{oil}} \cdot V_{\text{oil displaced}} \cdot g = 800 \cdot (0.01 \cdot h_{\text{oil}}) \cdot 9.8 = 78.4 \cdot h_{\text{oil}} , \text{N} ] 3. Solve for ( h_{\text{glycerol}} ) and ( h_{\text{oil}} ) Combine the forces: [ F_{\text{cube}} = F_{\text{glycerol}} + F_{\text{oil}} ] Substitute: [ 8.82 = 123.48 \cdot h_{\text{glycerol}} + 78.4 \cdot h_{\text{oil}} ] Since ( h_{\text{glycerol}} + h_{\text{oil}} = 0.1 ), we substitute ( h_{\text{oil}} = 0.1 - h_{\text{glycerol}} ): [ 8.82 = 123.48 \cdot h_{\text{glycerol}} + 78.4 \cdot (0.1 - h_{\text{glycerol}}) ] Simplify: [ 8.82 = 123.48 \cdot h_{\text{glycerol}} + 7.84 - 78.4 \cdot h_{\text{glycerol}} ] [ 8.82 - 7.84 = (123.48 - 78.4) \cdot h_{\text{glycerol}} ] [ 0.98 = 45.08 \cdot h_{\text{glycerol}} ] Solve for ( h_{\text{glycerol}} ): [ h_{\text{glycerol}} = \frac{0.98}{45.08} \approx 0.0217 , \text{m} = 2.17 , \text{cm}. ] Now, calculate ( h_{\text{oil}} ): [ h_{\text{oil}} = 0.1 - h_{\text{glycerol}} = 0.1 - 0.0217 = 0.0783 , \text{m} = 7.83 , \text{cm}. ] 4. Determine the Distance ( x ) The portion of the cube above the glycerol surface is: [ x = h_{\text{oil}} = 7.83 , \text{cm}. ] Final Answer The new distance ( x ) is 7.83 cm. Compared to the earlier result (( x = 2.85 , \text{cm} )), this confirms that ( x ) increases, aligning with the qualitative prediction in part (b). The addition of olive oil raises the cube further out of the glycerol due to its lower density.
xiaxie8 发表于 2025-03-23 01:17 今天发现，ChatGPT免费版可以正确解答这道AP Physics 1流体力学文字题3小问里面的第1问也就是Part a了。它依然无法理解Part b和c，但是哪天它理解了Part b和c，就意味着它能够在完全没有图片信息的情况下完整构建真实场景，我这个tutor100%将失业。这一天也许3年才到来，也许半年就到来。 You fill a 20-cm-tall container with glycerol so that the glycerol (density 1260 kg/m3) reaches the 10-cm mark. You place an oak cube (density 900 kg/m3) into the container. Each side of the cube is 10 cm. (a) What is the distance x from the upper face of the cube to the glycerol surface assuming the cube is in an upright position? (b) Make a prediction using qualitative reasoning (without mathematics) about what will happen to the cube in (a) if you now add olive oil (density 800 kg/m3) to the container until it is completely full. Will the distance x decrease, increase, or stay the same? (c) Determine the new distance x using mathematics (physics laws) and compare your result with the prediction in (b). 最新更新：今天网友们告知Copilot和ChatGPT o1收费版已经能够正确解答这题所有步骤！大家可能不知道这题(b)(c)两部分有多么困难。昨天我辅导TJ的2名AP Physics 1学生基本流体力学，她们Digital SAT数学都是790或800分，在我引入阿基米德原理以后，她们都能正确解出(a), 但她们都画不出(b)(c)的示意图，完全搞不懂受力分析。当我告诉她们怎么画图怎么做受力分析怎么解方程以后，她们花了30分钟才弄明白。而Copilot和ChatGPT o1已经打败几乎所有Digital SAT数学接近满分学生。而我也已经接近失业。
You fill a 20-cm-tall container with glycerol so that the glycerol (density 1260 kg/m3) reaches the 10-cm mark. You place an oak cube (density 900 kg/m3) into the container. Each side of the cube is 10 cm. (a) What is the distance x from the upper face of the cube to the glycerol surface assuming the cube is in an upright position? (b) Make a prediction using qualitative reasoning (without mathematics) about what will happen to the cube in (a) if you now add olive oil (density 800 kg/m3) to the container until it is completely full. Will the distance x decrease, increase, or stay the same? (c) Determine the new distance x using mathematics (physics laws) and compare your result with the prediction in (b).
最新更新：今天网友们告知Copilot和ChatGPT o1收费版已经能够正确解答这题所有步骤！大家可能不知道这题(b)(c)两部分有多么困难。昨天我辅导TJ的2名AP Physics 1学生基本流体力学，她们Digital SAT数学都是790或800分，在我引入阿基米德原理以后，她们都能正确解出(a), 但她们都画不出(b)(c)的示意图，完全搞不懂受力分析。当我告诉她们怎么画图怎么做受力分析怎么解方程以后，她们花了30分钟才弄明白。而Copilot和ChatGPT o1已经打败几乎所有Digital SAT数学接近满分学生。而我也已经接近失业。
Final Answer for (a) The distance ( x ) from the upper face of the cube to the glycerol surface is 2.85 cm.
(b) Here''s what would happen qualitatively: When you add olive oil (density ( 800 , \text{kg/m}^3 )) to the container, it will form a layer above the glycerol (density ( 1260 , \text{kg/m}^3 )) because olive oil is less dense and cannot mix with glycerol easily. The oak cube, which floats partly submerged due to glycerol''s buoyant force, will now interact with both the glycerol and olive oil. The olive oil provides a weaker buoyant force compared to glycerol, as its density is less than both glycerol and the oak cube. This means the glycerol will continue to bear most of the load of the cube, but the addition of the olive oil on top will exert an additional, albeit smaller, buoyant force. As a result, the cube will float higher overall, causing the distance ( x ) (from the cube''s upper face to the glycerol surface) to increase. The olive oil essentially reduces how much of the cube remains submerged in the glycerol, lifting the cube slightly farther above the glycerol''s surface. In short: The distance ( x ) will increase. Let me know if you''d like further clarification!
(c) Determine the new distance x using mathematics (physics laws) and compare your result with the prediction in (b).
To calculate the new distance ( x ), we consider the combined buoyant forces from glycerol and olive oil acting on the oak cube. Here''s a step-by-step breakdown:
1. Buoyant Force Analysis The cube now displaces glycerol and olive oil. The submerged height in glycerol is denoted ( h_{\text{glycerol}} ), and the submerged height in olive oil is ( h_{\text{oil}} ). Together: [ h_{\text{glycerol}} + h_{\text{oil}} = 0.1 , \text{m} , ( \text{total height of the cube} ) ]
2. Equilibrium Condition At equilibrium, the buoyant force equals the weight of the cube. The buoyant force is the sum of the forces due to glycerol and olive oil: [ F_{\text{buoyant}} = F_{\text{glycerol}} + F_{\text{oil}} ] The weight of the cube is: [ F_{\text{cube}} = \rho_{\text{oak}} \cdot V_{\text{cube}} \cdot g = 8.82 , \text{N} , (\text{as calculated earlier}). ] For glycerol: [ F_{\text{glycerol}} = \rho_{\text{glycerol}} \cdot V_{\text{glycerol displaced}} \cdot g ] The displaced glycerol volume is: [ V_{\text{glycerol displaced}} = \text{base area of cube} \cdot h_{\text{glycerol}} = 0.01 \cdot h_{\text{glycerol}} , \text{m}^3 ] So: [ F_{\text{glycerol}} = 1260 \cdot (0.01 \cdot h_{\text{glycerol}}) \cdot 9.8 = 123.48 \cdot h_{\text{glycerol}} , \text{N} ] For olive oil: [ F_{\text{oil}} = \rho_{\text{oil}} \cdot V_{\text{oil displaced}} \cdot g = 800 \cdot (0.01 \cdot h_{\text{oil}}) \cdot 9.8 = 78.4 \cdot h_{\text{oil}} , \text{N} ]
3. Solve for ( h_{\text{glycerol}} ) and ( h_{\text{oil}} ) Combine the forces: [ F_{\text{cube}} = F_{\text{glycerol}} + F_{\text{oil}} ] Substitute: [ 8.82 = 123.48 \cdot h_{\text{glycerol}} + 78.4 \cdot h_{\text{oil}} ] Since ( h_{\text{glycerol}} + h_{\text{oil}} = 0.1 ), we substitute ( h_{\text{oil}} = 0.1 - h_{\text{glycerol}} ): [ 8.82 = 123.48 \cdot h_{\text{glycerol}} + 78.4 \cdot (0.1 - h_{\text{glycerol}}) ] Simplify: [ 8.82 = 123.48 \cdot h_{\text{glycerol}} + 7.84 - 78.4 \cdot h_{\text{glycerol}} ] [ 8.82 - 7.84 = (123.48 - 78.4) \cdot h_{\text{glycerol}} ] [ 0.98 = 45.08 \cdot h_{\text{glycerol}} ] Solve for ( h_{\text{glycerol}} ): [ h_{\text{glycerol}} = \frac{0.98}{45.08} \approx 0.0217 , \text{m} = 2.17 , \text{cm}. ] Now, calculate ( h_{\text{oil}} ): [ h_{\text{oil}} = 0.1 - h_{\text{glycerol}} = 0.1 - 0.0217 = 0.0783 , \text{m} = 7.83 , \text{cm}. ]
4. Determine the Distance ( x ) The portion of the cube above the glycerol surface is: [ x = h_{\text{oil}} = 7.83 , \text{cm}. ]
Final Answer The new distance ( x ) is 7.83 cm. Compared to the earlier result (( x = 2.85 , \text{cm} )), this confirms that ( x ) increases, aligning with the qualitative prediction in part (b). The addition of olive oil raises the cube further out of the glycerol due to its lower density.
以前是学好数理化，走遍世界都不怕； 现在好了，打开抖音看世界，娱乐至上不用怕。
免费的 还是 收费的版本？
我只有免费的ChatGPT。
那个AP Physics 1简单流体力学题的（a), ChatGPT, DeepSeek, Gemini, Copilot都可以正确计算。
但是(b)(c)需要机器能够像人一样把文字转化为图形，大部分机器暂时做不到，但前面层主的Copilot做到了。Copilot目前名列第一！
Copilot完全正确！！！
我已经完蛋了。
chatgpt o1 也能算对 我觉得 AI 对 tutor 行业的打击还是很厉害的。 不过你算资深了， 要打击也是最后一轮了。 你也许都退休了。
过去几个月几个主要AI绝对是指数式爆发。
从解AP Physics 1纯文字题水平看，和Copilot相比，TJ那些Digital SAT考790-800的学生都是白痴。
顶尖程序员将主宰世界。
而我只是蹩脚程序员。
它们互相学习。
很快它们就不需要人类了。
多谢您的最新信息！
为什么付费的ChatGPT O1水平远高于免费的ChatGPT？它们是不同模型？
谢谢！
我觉得八成是预训练过类似的题目。现实世界还是复杂太多了
牛啊。
你这个基本是马克思资本论的结论
照这个发展速度，除非是已经快要退休的人，估计我们这一代的人就得赶上，何解？
我用的Gemini :
好的，我来帮您解答这道物理题。
**(a) 计算橡木块上表面到甘油表面的距离 x**
1. **橡木块的体积：** * $V_{cube} = (10 \, \text{cm})^3 = 1000 \, \text{cm}^3 = 0.001 \, \text{m}^3$
2. **橡木块的质量：** * $m_{cube} = \rho_{oak} \cdot V_{cube} = 900 \, \text{kg/m}^3 \cdot 0.001 \, \text{m}^3 = 0.9 \, \text{kg}$
3. **橡木块在甘油中漂浮，浮力等于重力：** * $F_{buoyant} = F_{gravity}$ * $\rho_{glycerol} \cdot V_{submerged} \cdot g = m_{cube} \cdot g$
4. **计算浸入甘油中的体积：** * $V_{submerged} = \frac{m_{cube}}{\rho_{glycerol}} = \frac{0.9 \, \text{kg}}{1260 \, \text{kg/m}^3} = 0.000714 \, \text{m}^3 = 714 \, \text{cm}^3$
5. **计算浸入甘油中的高度：** * $h_{submerged} = \frac{V_{submerged}}{A_{cube}} = \frac{714 \, \text{cm}^3}{(10 \, \text{cm})^2} = 7.14 \, \text{cm}$
6. **计算 x：** * $x = 10 \, \text{cm} - 7.14 \, \text{cm} = 2.86 \, \text{cm}$
**(b) 定性预测：加入橄榄油后的情况**
* 由于橄榄油的密度小于甘油，它会漂浮在甘油之上。 * 橄榄油会增加液体的总高度，同时也会对橡木块产生额外的浮力。 * 橡木块受到的总浮力会增加，因此浸入甘油的体积会减少。 * 因此，x 会增加。
**(c) 计算加入橄榄油后的 x**
1. **橄榄油的高度：** * $h_{oil} = 20 \, \text{cm} - 10 \, \text{cm} = 10 \, \text{cm}$
2. **橡木块受到的总浮力：** * $F_{buoyant} = F_{buoyant, glycerol} + F_{buoyant, oil}$ * $F_{buoyant} = \rho_{glycerol} \cdot V_{glycerol, submerged} \cdot g + \rho_{oil} \cdot V_{oil, submerged} \cdot g$ * $F_{buoyant} = m_{cube} \cdot g$
3. **设橡木块在甘油中的浸没高度为 h，则在橄榄油中的浸没高度为 (10 - h)：** * $\rho_{glycerol} \cdot (h \cdot 100 \, \text{cm}^2) \cdot g + \rho_{oil} \cdot ((10 - h) \cdot 100 \, \text{cm}^2) \cdot g = m_{cube} \cdot g$
4. **代入数值并求解 h：** * $1260 \cdot h + 800 \cdot (10 - h) = 900$ * $1260h + 8000 - 800h = 900$ * $460h = -7100$ * $h = 5 \, \text{cm}$
5. **计算 x：** * $x = 10 \, \text{cm} - 5 \, \text{cm} = 5 \, \text{cm}$
* **结论：** * 加入橄榄油后，x 从 2.86 cm 增加 5 cm到7.86cm，与预测一致。
您明白了吗？
(a) 首先计算橡木立方体的重量。立方体的体积为 \(0.1 \, \text{m} \times 0.1 \, \text{m} \times 0.1 \, \text{m} = 0.001 \, \text{m}^3\)，质量为 \(900 \, \text{kg/m}^3 \times 0.001 \, \text{m}^3 = 0.9 \, \text{kg}\)。根据浮力原理，排开的甘油重量等于立方体重量，得到排开体积 \(V = \frac{0.9 \, \text{kg}}{1260 \, \text{kg/m}^3} \approx 0.000714 \, \text{m}^3\)。立方体底面积为 \(0.1 \, \text{m} \times 0.1 \, \text{m} = 0.01 \, \text{m}^2\)，因此浸入深度 \(h = \frac{0.000714 \, \text{m}^3}{0.01 \, \text{m}^2} \approx 0.0714 \, \text{m} = 7.14 \, \text{cm}\)。由于原甘油液面为10 cm，立方体浸入7.14 cm后，顶部到液面的距离 \(x = 10 \, \text{cm} - 7.14 \, \text{cm} \approx 2.86 \, \text{cm}\)。
(b) 当加入橄榄油后，由于橄榄油密度小于橡木，无法提供浮力，立方体依然仅浸入甘油中。此时液面升至20 cm，立方体顶部到新液面的距离 \(x\) 由原来的2.86 cm变为20 cm - 10 cm = 10 cm，因此 \(x\) 增加。
(c) 数学计算确认浸入甘油深度不变，立方体顶部到橄榄油液面的距离为20 cm - 10 cm = 10 cm，与预测一致，\(x\) 增加。
### 最终答案 (a) \boxed{2.86 \text{ cm}}
(b) 距离x会增加。
(c) 新的距离x为\boxed{10 \text{ cm}}，与预测一致，x增加。
确实进步神速啊。
在一个正方体牢笼里有一只雄鸟，每过一秒钟，该雄鸟都会随机选择该正方体的各个面之一飞出去，牢笼底部是钢筋地板．如果从上面飞出去会遇到新的正方体牢笼就会重新选择飞行，且上面的牢笼可以无限多个，各个侧面飞出去就可以获得自由，那么在 2023 秒后，该雄鸟获得自由的概率是多少？
Xiaxie 网友能做一下这道题吗？ AI 给出了不同的答案。
您这次Gemini完全正确。
DeepSeek这次b和c都是错误的。
网上找不到答案，因为是AP Physics 1课本上面的偶数题，而只有奇数题才有答案。
这道题是条件概率，并不难。但文字方面有歧义：现有牢笼底部是钢筋地板，那肯定飞不出去，也不能重新选择飞行，而现有牢笼顶部正好对着上面一个牢笼底部钢筋，肯定飞不出去，但有重新选择飞行的一次机会。这种条件下，上面牢笼数量无论是一个还是无穷个，没有任何区别！
难道没钱去印度雇大学生把历史题目都喂几遍？
那您的意思是，获得自由的概率是趋近于1，如果时间/ 秒数足够大？
DS给出的答案是1减去 1/3的2023 次方。
DS性能不稳定，问题可能处在语言的歧义上面。我刚问DS它给出了不同答案，并且是明显错误答案：
### 问题理解
我们有一个正方体牢笼，里面有一只雄鸟。每过一秒钟，这只鸟会随机选择一个面飞出去。牢笼的底部是钢筋地板，意味着如果鸟选择向下飞，它会被困住。如果鸟选择向上飞，它会进入一个新的正方体牢笼，这个过程可以无限重复。如果鸟选择任何一个侧面飞出去，它就能获得自由。问题是：在2023秒后，这只鸟获得自由的概率是多少？
### 分析步骤
1. **确定每个面的选择概率**： - 正方体有6个面：上、下、前、后、左、右。 - 鸟随机选择一个面，每个面被选择的概率都是1/6。
2. **定义状态**： - 设 \( P(n) \) 为在第 \( n \) 秒时鸟获得自由的概率。 - 我们需要计算 \( P(2023) \)。
3. **建立递推关系**： - 如果鸟选择向上飞（概率1/6），它会进入一个新的牢笼，相当于重新开始，概率仍然是 \( P(n-1) \)。 - 如果鸟选择向下飞（概率1/6），它会被困住，无法获得自由，概率为0。 - 如果鸟选择任何一个侧面飞出去（概率4/6），它立即获得自由，概率为1。
因此，递推关系为： \[ P(n) = \frac{1}{6}P(n-1) + \frac{4}{6} \times 1 + \frac{1}{6} \times 0 \] 简化后： \[ P(n) = \frac{1}{6}P(n-1) + \frac{2}{3} \]
4. **求解递推关系**： 这是一个线性非齐次递推关系。我们可以先求其通解。
- **齐次方程**： \[ P_h(n) = A \left( \frac{1}{6} \right)^n \] - **特解**： 假设特解为常数 \( P_p(n) = C \)，代入递推关系： \[ C = \frac{1}{6}C + \frac{2}{3} \] 解得： \[ C - \frac{1}{6}C = \frac{2}{3} \Rightarrow \frac{5}{6}C = \frac{2}{3} \Rightarrow C = \frac{4}{5} \] - **通解**： \[ P(n) = A \left( \frac{1}{6} \right)^n + \frac{4}{5} \] - **确定常数 \( A \)**： 初始条件：当 \( n = 0 \) 时，鸟还未开始飞，获得自由的概率为0。 \[ 0 = A \left( \frac{1}{6} \right)^0 + \frac{4}{5} \Rightarrow A + \frac{4}{5} = 0 \Rightarrow A = -\frac{4}{5} \] - **最终解**： \[ P(n) = -\frac{4}{5} \left( \frac{1}{6} \right)^n + \frac{4}{5} = \frac{4}{5} \left( 1 - \left( \frac{1}{6} \right)^n \right) \]
5. **计算 \( P(2023) \)**： 由于 \( \left( \frac{1}{6} \right)^{2023} \) 是一个非常小的数，接近于0，因此： \[ P(2023) \approx \frac{4}{5} \times 1 = \frac{4}{5} = 0.8 \] 即，鸟在2023秒后获得自由的概率约为80%。
### 最终答案
在2023秒后，该雄鸟获得自由的概率约为80%。
在给定无限次尝试的情况下，雄鸟逃脱的概率确实趋近于1，即几乎必然发生。这是因为每次向上飞会进入一个完全相同的新牢笼，使得雄鸟始终有机会通过侧面逃脱。
这样理解是不是就简单多了？
您的DS基于上方一个牢笼的底部是可以进但不能出的，递归计算结果是完全正确的！雄鸟尝试越多次数肯定获得自由概率越接近1。
我对DS的印象是：数学和物理非常强，但汉语语言有一定歧义。
让DS 写诗，它是非常擅长的，只是写得意象杂乱，不知所云，非常炫技，一看就是DS写的，辨识度极高。
如果把题目改成尝试5秒后，获得自由的概率是多少，您再试试，看AI能否做对？
DS可能真的会做。我把题目改成5秒后，它依然算对。
Gemini 给出的答案是：
* 设 P(n) 为 n 秒后雄鸟获得自由的概率。
* 第一秒，雄鸟有 2/3 的概率获得自由，1/6 的概率选择顶部，1/6 的概率选择底部。
* 如果雄鸟选择顶部，它会重新开始，此时获得自由的概率为 P(n-1)。
* 因此，我们可以得到递推公式： * P(n) = 2/3 + (1/6) \* P(n-1)
* 初始条件：
* P(0) = 0
* 根据递推公式，我们可以计算出 P(1)、P(2)、P(3)、P(4)、P(5) 的值。
* P(1) = 2/3 * P(2) = 2/3 + (1/6) \* (2/3) = 7/9 * P(3) = 2/3 + (1/6) \* (7/9) = 47/54 * P(4) = 2/3 + (1/6) \* (47/54) = 293/324 * P(5) = 2/3 + (1/6) \* (293/324) = 1799/1944
二者的递推公式不同， 是不是对题目文字的理解不同？ 按照xiaxie 网友的理解，DS是正确的。题目本身的文字叙述确实有歧义。
是的！
这道题是比较简单的条件概率题。
AI写诗都是如此。
AI写歌听起来很华丽，但过后人毫无印象。
那更难一些，不能简单递归，只能一步步地计算。
这道题5秒不等于5步，Gemini的递归似乎更正确！
(1) 求该系学生甲收到李老师或张老师所发活动通知信息的概率;
(II) 求使P（X=m ）取得最大值的整数m 。
这道题没有歧义，我要试试AI 们。
为何5秒不等于5步呢？ Gemini 也是给出 递推公式，然后代入n=5 ，得到1799/1944的结果。
我也感觉这个结果比242/243更make sense.
说得好， 这也是我对人工智能最怕的地方
我搜了，好像是有的啊
什么是创新？ 如果创新的本质并不是人类想象的那么神圣 通过概率计算也能产生创新 那人类就彻底玩完了
如果网上原来有答案，那不能排除一些高级AI事先存储了答案。
但是昨天我拿这个流体力学题问ChatGPT免费版时，它只正确回答了a，而对b和c瞎说x不变，这100%表明它事先根本就没有存储答案。
又训练了