Sure! Let's solve the problem step-by-step:
### Given Data:
1. Radius of each cylinder: [tex]\( r = 0.25 \, \text{cm} = 0.25 \times 10^{-2} \, \text{m} = 0.0025 \, \text{m} \)[/tex]
2. Compressive force: [tex]\( F = 6500 \, \text{N} \)[/tex]
3. Young's Modulus for Copper (Y_copper): [tex]\( Y_{\text{copper}} = 1.1 \times 10^{11} \, \text{Pa} \)[/tex]
4. Young's Modulus for Brass (Y_brass): [tex]\( Y_{\text{brass}} = 9.0 \times 10^{10} \, \text{Pa} \)[/tex]
5. Original length of the Copper cylinder (L0_copper): [tex]\( L0_{\text{copper}} = 0.03 \, \text{m} \)[/tex]
6. Original length of the Brass cylinder (L0_brass): [tex]\( L0_{\text{brass}} = 0.03 \, \text{m} \)[/tex]
### Step-by-Step Solution:
1. Calculate the cross-sectional area (A) of each cylinder:
[tex]\[
A = \pi r^2 = \pi (0.0025 \, \text{m})^2 = \pi \times 6.25 \times 10^{-6} \, \text{m}^2 = 1.963495408 \times 10^{-5} \, \text{m}^2.
\][/tex]
2. Calculate the deformation (ΔL) for the copper cylinder:
[tex]\[
\Delta L_{\text{copper}} = \frac{F \cdot L0_{\text{copper}}}{Y_{\text{copper}} \cdot A}
\][/tex]
Substitute the values:
[tex]\[
\Delta L_{\text{copper}} = \frac{6500 \, \text{N} \times 0.03 \, \text{m}}{1.1 \times 10^{11} \, \text{Pa} \times 1.963495408 \times 10^{-5} \, \text{m}^2}.
\][/tex]
[tex]\[
\Delta L_{\text{copper}} \approx 9.028425862667518 \times 10^{-5} \, \text{m} = 0.000090284 \, \text{m}.
\][/tex]
3. Calculate the deformation (ΔL) for the brass cylinder:
[tex]\[
\Delta L_{\text{brass}} = \frac{F \cdot L0_{\text{brass}}}{Y_{\text{brass}} \cdot A}
\][/tex]
Substitute the values:
[tex]\[
\Delta L_{\text{brass}} = \frac{6500 \, \text{N} \times 0.03 \, \text{m}}{9.0 \times 10^{10} \, \text{Pa} \times 1.963495408 \times 10^{-5} \, \text{m}^2}.
\][/tex]
[tex]\[
\Delta L_{\text{brass}} \approx 0.00011034742721038078 \, \text{m} = 0.000110347 \, \text{m}.
\][/tex]
4. Calculate the total decrease in length of the stack:
[tex]\[
\Delta L_{\text{total}} = \Delta L_{\text{copper}} + \Delta L_{\text{brass}}
\][/tex]
[tex]\[
\Delta L_{\text{total}} = 0.000090284 \, \text{m} + 0.000110347 \, \text{m}.
\][/tex]
[tex]\[
\Delta L_{\text{total}} \approx 0.00020063168583705595 \, \text{m} = 0.000200632 \, \text{m}.
\][/tex]
### Conclusion:
The amount by which the length of the stack decreases is approximately [tex]\(0.000200632 \, \text{m}\)[/tex] or [tex]\(0.20063 \, \text{mm}\)[/tex].
