In the figure, cube A is 30 cm along each edge and weighs 45 kg. It is attached to the square prism B which is 15cm * 15 deg * cm * 240cm and weighs 480kg / (m ^ 3) What length of B will project above the liquid surface? Specific gravity of liquid is 1.5.

Let's break down the problem and solve it step by step.

### Step 1: Analyze the components of the system

- **Cube A**: 
  - Edge length = 30 cm (or 0.30 m) 
  - Volume of cube \(V_A\) = \(0.30^3\) cubic meters
  - Weight of cube A = 45 kg

- **Prism B**: 
  - Base area = \(15 \, \text{cm} \times 15 \, \text{cm} = 0.15 \, \text{m} \times 0.15 \, \text{m} = 0.0225 \, \text{m}^2\)
  - Height = 240 cm = 2.4 m
  - Density of prism B = 480 kg/m³
  - Volume of prism \(V_B\) = Base area × Height = \(0.0225 \, \text{m}^2 \times 2.4 \, \text{m} = 0.054 \, \text{m}^3\)

  Therefore, mass of prism B is:
  \[
  \text{Mass of B} = \text{Density} \times \text{Volume} = 480 \, \text{kg/m}^3 \times 0.054 \, \text{m}^3 = 25.92 \, \text{kg}
  \]

### Step 2: Calculate total weight of the system
The total weight is the sum of the weight of cube A and the mass of prism B.

\[
\text{Total weight} = 45 \, \text{kg} + 25.92 \, \text{kg} = 70.92 \, \text{kg}
\]

### Step 3: Analyze the buoyancy forces

The specific gravity of the liquid is 1.5, which means the density of the liquid is:

\[
\text{Density of liquid} = 1.5 \times 1000 \, \text{kg/m}^3 = 1500 \, \text{kg/m}^3
\]

The buoyant force acting on the system is equal to the weight of the liquid displaced, given by Archimedes' principle:

\[
\text{Buoyant force} = \text{Weight of displaced liquid} = \text{Volume submerged} \times \text{Density of liquid} \times g
\]

Where \(g = 9.81 \, \text{m/s}^2\).

### Step 4: Submerged volume and equilibrium condition

At equilibrium, the weight of the system is balanced by the buoyant force. Thus:

\[
70.92 \, \text{kg} \times 9.81 \, \text{m/s}^2 = (\text{Volume submerged}) \times 1500 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2
\]

Simplifying:

\[
70.92 = (\text{Volume submerged}) \times 1500
\]

\[
\text{Volume submerged} = \frac{70.92}{1500} = 0.04728 \, \text{m}^3
\]

### Step 5: Calculate the submerged length of prism B

The total submerged volume comes from both cube A and part of prism B. The volume of cube A is:

\[
V_A = 0.30^3 = 0.027 \, \text{m}^3
\]

Let the submerged length of prism B be \(h\). The volume submerged for prism B is:

\[
\text{Volume of B submerged} = 0.0225 \, \text{m}^2 \times h
\]

Therefore, the total submerged volume is:

\[
0.027 + 0.0225 \times h = 0.04728
\]

Solving for \(h\):

\[
0.0225 \times h = 0.04728 - 0.027 = 0.02028
\]

\[
h = \frac{0.02028}{0.0225} = 0.901 \, \text{m}
\]

### Step 6: Calculate the portion of prism B above the liquid

The total height of prism B is 2.4 m. Therefore, the length of prism B projecting above the liquid surface is:

\[
2.4 \, \text{m} - 0.901 \, \text{m} = 1.499 \, \text{m}
\]

### Final Answer:

The length of prism B projecting above the liquid surface is approximately **1.499 meters**.

---

Do you want any further explanation or details on this solution?

Here are 5 related questions to explore further:
1. How does increasing the weight of cube A affect the submerged length of prism B?
2. What happens if the specific gravity of the liquid changes to 1.2?
3. How would the result change if the density of prism B were halved?
4. How does the system behave if cube A were made of a material with a different density?
5. What if prism B was shortened to 1.5 m while keeping other factors the same?

### Tip:
Always check that the units you use are consistent throughout the problem (e.g., converting cm to m when needed).

In the figure, cube A is 30 cm along each edge and weighs 45 kg. It is attached to the square prism B which is 15 cm * 15 cm * 240 cm and weighs 480 kg/m^3. What length of B will project above the liquid surface? Specific gravity of liquid is 1.5.

Solve a problem involving a 30 cm cube attached to a 15 cm by 240 cm prism submerged in a liquid with specific gravity 1.5. Learn how to calculate the length of the prism projecting above the liquid using Archimedes' principle and key formulas like volume and buoyant force. Suitable for high school students in Grades 10-12.

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