Question
A steel pipe 100 cm long has an outside diameter of 2 cm and an inside diameter of 1.8 cm. If the density of the steel is 7.8 grams per cm3, what is the mass of the pipe to the nearest gram?
Answers
GPT-4o mini
To calculate the mass of the steel pipe, we first need to find the volume of the steel in the pipe. The volume of the steel can be calculated by taking the difference between the volume of the outer cylinder (the entire pipe) and the volume of the inner cylinder (the hollow part).
1. Calculate the outer radius and inner radius:
- The outer radius \( r_{\text{outer}} = \frac{\text{outer diameter}}{2} = \frac{2 \text{ cm}}{2} = 1 \text{ cm} \)
- The inner radius \( r_{\text{inner}} = \frac{\text{inner diameter}}{2} = \frac{1.8 \text{ cm}}{2} = 0.9 \text{ cm} \)
2. The height (length) of the pipe \( h = 100 \text{ cm} \).
3. Calculate the volume of the outer cylinder (V_outer):
\[
V_{\text{outer}} = \pi r_{\text{outer}}^2 h = \pi (1 \text{ cm})^2 (100 \text{ cm}) = \pi (1) (100) = 100\pi \text{ cm}^3
\]
4. Calculate the volume of the inner cylinder (V_inner):
\[
V_{\text{inner}} = \pi r_{\text{inner}}^2 h = \pi (0.9 \text{ cm})^2 (100 \text{ cm}) = \pi (0.81) (100) = 81\pi \text{ cm}^3
\]
5. The volume of the steel in the pipe \( V_{\text{steel}} \) is:
\[
V_{\text{steel}} = V_{\text{outer}} - V_{\text{inner}} = 100\pi \text{ cm}^3 - 81\pi \text{ cm}^3 = 19\pi \text{ cm}^3
\]
6. Now calculate the mass of the steel using the density \( \rho \):
- Density \( \rho = 7.8 \text{ g/cm}^3 \)
\[
\text{mass} = \rho \times V_{\text{steel}} = 7.8 \text{ g/cm}^3 \times 19\pi \text{ cm}^3
\]
7. Calculate \( \pi \approx 3.14 \):
\[
V_{\text{steel}} \approx 19 \times 3.14 \approx 59.66 \text{ cm}^3
\]
Thus,
\[
\text{mass} \approx 7.8 \text{ g/cm}^3 \times 59.66 \text{ cm}^3 \approx 465.428 \text{ grams}
\]
8. Rounding to the nearest gram, the mass of the pipe is approximately:
\[
\boxed{465} \text{ grams}
\]
1. Calculate the outer radius and inner radius:
- The outer radius \( r_{\text{outer}} = \frac{\text{outer diameter}}{2} = \frac{2 \text{ cm}}{2} = 1 \text{ cm} \)
- The inner radius \( r_{\text{inner}} = \frac{\text{inner diameter}}{2} = \frac{1.8 \text{ cm}}{2} = 0.9 \text{ cm} \)
2. The height (length) of the pipe \( h = 100 \text{ cm} \).
3. Calculate the volume of the outer cylinder (V_outer):
\[
V_{\text{outer}} = \pi r_{\text{outer}}^2 h = \pi (1 \text{ cm})^2 (100 \text{ cm}) = \pi (1) (100) = 100\pi \text{ cm}^3
\]
4. Calculate the volume of the inner cylinder (V_inner):
\[
V_{\text{inner}} = \pi r_{\text{inner}}^2 h = \pi (0.9 \text{ cm})^2 (100 \text{ cm}) = \pi (0.81) (100) = 81\pi \text{ cm}^3
\]
5. The volume of the steel in the pipe \( V_{\text{steel}} \) is:
\[
V_{\text{steel}} = V_{\text{outer}} - V_{\text{inner}} = 100\pi \text{ cm}^3 - 81\pi \text{ cm}^3 = 19\pi \text{ cm}^3
\]
6. Now calculate the mass of the steel using the density \( \rho \):
- Density \( \rho = 7.8 \text{ g/cm}^3 \)
\[
\text{mass} = \rho \times V_{\text{steel}} = 7.8 \text{ g/cm}^3 \times 19\pi \text{ cm}^3
\]
7. Calculate \( \pi \approx 3.14 \):
\[
V_{\text{steel}} \approx 19 \times 3.14 \approx 59.66 \text{ cm}^3
\]
Thus,
\[
\text{mass} \approx 7.8 \text{ g/cm}^3 \times 59.66 \text{ cm}^3 \approx 465.428 \text{ grams}
\]
8. Rounding to the nearest gram, the mass of the pipe is approximately:
\[
\boxed{465} \text{ grams}
\]