请以资深中国物理教师的身份，解析图片上习题：质量分布均匀的实心正方体A、B置于水平桌面上，如图甲。将B沿水平方向截取高为h的柱体，并将该柱体叠放在A上，A、B剩余部分对桌面的压强p随截取高度h的变化关系如图乙，则B的密度为 ，A的重力为 N（g取10N/kg）---**Image Description:**
The image contains two distinct parts labeled "甲" and "乙".
**Part 甲:**
This part shows a diagram depicting two rectangular blocks, labeled "A" and "B", resting side-by-side on a horizontal surface represented by a hatched line. Block A is larger in both height and width than block B.
**Part 乙:**
This part shows a two-dimensional line chart.
- **Type:** Line chart.
- **Coordinate Axes:**
- The vertical axis is labeled "$p / \times 10^2 \text{Pa}$", representing pressure scaled by $10^2$ Pa. The axis origin is at 0. Major tick marks are shown at 0, 6, and 8. An intermediate tick mark is visible at 2 on the horizontal axis.
- The horizontal axis is labeled "$h / \text{cm}$", representing height or depth in centimeters. The axis origin is at 0 and increases to the right. Major tick marks are shown at 0, 2, and 10.
- **Data Series:**
- There are two lines labeled "A" and "B".
- **Line A:** A gray straight line increasing from a point on the vertical axis (approximately $p=4 \times 10^2$ Pa at $h=0$) to a point at $(h=10 \text{ cm}, p=8 \times 10^2 \text{ Pa})$. It passes through a point near $(h=2 \text{ cm}, p \approx 5 \times 10^2 \text{ Pa})$. Dashed lines indicate the coordinates of the point $(10, 8)$ on the graph.
- **Line B:** A gray straight line decreasing from a point on the vertical axis ($p=6 \times 10^2 \text{ Pa}$ at $h=0$) to a point on the horizontal axis ($p=0 \text{ Pa}$ at $h=10 \text{ cm}$). It passes through a point near $(h=2 \text{ cm}, p \approx 5 \times 10^2 \text{ Pa})$, intersecting line A. Dashed lines indicate the $h=2 \text{ cm}$ coordinate of the intersection point.
- **Labels and Annotations:** The lines are labeled "A" and "B". The axes are labeled with the physical quantity and unit/scaling. The parts of the image are labeled "甲" and "乙".
There is no question stem, options, or other explanatory text present in the image.
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Let's analyze this pressure problem step by step. We have two uniform solid cubes A and B on a table. When we cut a cylinder of height h from cube B and place it on cube A, the pressure on the table changes. From the graph, we can determine the density of B and the weight of A.
Now let's analyze the pressure-height graph. Line B shows the pressure of the remaining part of cube B, decreasing from 600 Pa to zero as we cut more material. When h equals 10 cm, the pressure becomes zero, meaning we've cut the entire cube B. Line A shows the pressure under cube A increasing as we add more material on top. The intersection point at h equals 2 cm and p equals 500 Pa will be important for our calculations.
Let's calculate the density of cube B. The pressure formula for the remaining part of B is p equals rho times g times L minus h, where L is the original height. From the graph, when h equals 10 cm, the pressure becomes zero, so the original height L must be 10 cm. At h equals zero, the pressure is 600 Pa. Substituting into our formula: 600 equals rho times 10 times 0.1, which gives us rho equals 600 kg per cubic meter.
Now let's find the weight of cube A. The pressure under A includes both A's weight and the weight of the cut material from B. The cut material has weight 60h Newtons per meter. From the graph, we know the pressure increases linearly from 400 Pa to 800 Pa. Using the two data points, we can solve for the base area of A as 0.015 square meters. Therefore, the weight of A is 400 times 0.015, which equals 6 Newtons.
Let's summarize our solution. By analyzing the pressure-height graph, we determined that cube B has an original height of 10 cm. Using the pressure formula for the remaining part of B, we calculated its density as 600 kg per cubic meter. For cube A, we used the linear relationship between pressure and added height to find its base area, leading to a weight of 6 Newtons. Therefore, our final answers are: the density of B is 600 kg per cubic meter, and the weight of A is 6 Newtons.