Discover the answers you need at Westonci.ca, a dynamic Q&A platform where knowledge is shared freely by a community of experts. Our Q&A platform offers a seamless experience for finding reliable answers from experts in various disciplines. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
To answer the question, let's break it down into two parts:
### Part 1: Calculating Linear Expansivity of the Metal Rod
Given:
- Initial length of the rod [tex]\( L_0 = 40 \, \text{cm} \)[/tex]
- Final length of the rod [tex]\( L_1 = 40.05 \, \text{cm} \)[/tex]
- Initial temperature [tex]\( T_0 = 0^\circ \text{C} \)[/tex]
- Final temperature [tex]\( T_1 = 45^\circ \text{C} \)[/tex]
The change in length of the rod is:
[tex]\[ \Delta L = L_1 - L_0 = 40.05 \, \text{cm} - 40 \, \text{cm} = 0.05 \, \text{cm} \][/tex]
The change in temperature is:
[tex]\[ \Delta T = T_1 - T_0 = 45^\circ \text{C} - 0^\circ \text{C} = 45 \, \text{K} \][/tex]
Linear expansivity ([tex]\(\alpha\)[/tex]) is given by the formula:
[tex]\[ \alpha = \frac{\Delta L}{L_0 \cdot \Delta T} \][/tex]
Substituting the known values:
[tex]\[ \alpha = \frac{0.05 \, \text{cm}}{40 \, \text{cm} \cdot 45 \, \text{K}} \][/tex]
[tex]\[ \alpha = \frac{0.05}{40 \times 45} \][/tex]
[tex]\[ \alpha = \frac{0.05}{1800} \][/tex]
[tex]\[ \alpha = \frac{5 \times 10^{-2}}{1.8 \times 10^3} \][/tex]
[tex]\[ \alpha = \frac{5}{1800} \times 10^{-2} \][/tex]
[tex]\[ \alpha = \frac{1}{360} \times 10^{-2} \][/tex]
[tex]\[ \alpha = \frac{1}{3.6 \times 10^2} \times 10^{-2} \][/tex]
[tex]\[ \alpha = \frac{1}{3.6} \times 10^{-5} \][/tex]
[tex]\[ \alpha \approx 2.78 \times 10^{-5} \, \text{K}^{-1} \][/tex]
### Part 2: Calculating P.d Across the Terminal
Given:
- Electromotive force (E.M.F): [tex]\( \mathcal{E} = 12 \, \text{V} \)[/tex]
- Internal resistance: [tex]\( r = 2 \, \Omega \)[/tex]
- External resistance: [tex]\( R = 8 \, \Omega \)[/tex]
First, calculate the total resistance in the circuit:
[tex]\[ R_{\text{total}} = r + R = 2 \Omega + 8 \Omega = 10 \Omega \][/tex]
Next, use Ohm's Law to calculate the current ([tex]\(I\)[/tex]) flowing through the circuit:
[tex]\[ I = \frac{\mathcal{E}}{R_{\text{total}}} = \frac{12 \, \text{V}}{10 \, \Omega} = 1.2 \, \text{A} \][/tex]
Finally, calculate the potential difference (P.d) across the terminal using the formula:
[tex]\[ V_{\text{terminal}} = \mathcal{E} - I \cdot r = 12 \, \text{V} - 1.2 \, \text{A} \times 2 \, \Omega = 12 \, \text{V} - 2.4 \, \text{V} = 9.6 \, \text{V} \][/tex]
Thus, the P.d across the terminal is:
[tex]\[ \boxed{9.6 \, \text{V}} \][/tex]
The correct answer for the second problem is: B. [tex]\( 9.6 \, \text{V} \)[/tex]
### Part 1: Calculating Linear Expansivity of the Metal Rod
Given:
- Initial length of the rod [tex]\( L_0 = 40 \, \text{cm} \)[/tex]
- Final length of the rod [tex]\( L_1 = 40.05 \, \text{cm} \)[/tex]
- Initial temperature [tex]\( T_0 = 0^\circ \text{C} \)[/tex]
- Final temperature [tex]\( T_1 = 45^\circ \text{C} \)[/tex]
The change in length of the rod is:
[tex]\[ \Delta L = L_1 - L_0 = 40.05 \, \text{cm} - 40 \, \text{cm} = 0.05 \, \text{cm} \][/tex]
The change in temperature is:
[tex]\[ \Delta T = T_1 - T_0 = 45^\circ \text{C} - 0^\circ \text{C} = 45 \, \text{K} \][/tex]
Linear expansivity ([tex]\(\alpha\)[/tex]) is given by the formula:
[tex]\[ \alpha = \frac{\Delta L}{L_0 \cdot \Delta T} \][/tex]
Substituting the known values:
[tex]\[ \alpha = \frac{0.05 \, \text{cm}}{40 \, \text{cm} \cdot 45 \, \text{K}} \][/tex]
[tex]\[ \alpha = \frac{0.05}{40 \times 45} \][/tex]
[tex]\[ \alpha = \frac{0.05}{1800} \][/tex]
[tex]\[ \alpha = \frac{5 \times 10^{-2}}{1.8 \times 10^3} \][/tex]
[tex]\[ \alpha = \frac{5}{1800} \times 10^{-2} \][/tex]
[tex]\[ \alpha = \frac{1}{360} \times 10^{-2} \][/tex]
[tex]\[ \alpha = \frac{1}{3.6 \times 10^2} \times 10^{-2} \][/tex]
[tex]\[ \alpha = \frac{1}{3.6} \times 10^{-5} \][/tex]
[tex]\[ \alpha \approx 2.78 \times 10^{-5} \, \text{K}^{-1} \][/tex]
### Part 2: Calculating P.d Across the Terminal
Given:
- Electromotive force (E.M.F): [tex]\( \mathcal{E} = 12 \, \text{V} \)[/tex]
- Internal resistance: [tex]\( r = 2 \, \Omega \)[/tex]
- External resistance: [tex]\( R = 8 \, \Omega \)[/tex]
First, calculate the total resistance in the circuit:
[tex]\[ R_{\text{total}} = r + R = 2 \Omega + 8 \Omega = 10 \Omega \][/tex]
Next, use Ohm's Law to calculate the current ([tex]\(I\)[/tex]) flowing through the circuit:
[tex]\[ I = \frac{\mathcal{E}}{R_{\text{total}}} = \frac{12 \, \text{V}}{10 \, \Omega} = 1.2 \, \text{A} \][/tex]
Finally, calculate the potential difference (P.d) across the terminal using the formula:
[tex]\[ V_{\text{terminal}} = \mathcal{E} - I \cdot r = 12 \, \text{V} - 1.2 \, \text{A} \times 2 \, \Omega = 12 \, \text{V} - 2.4 \, \text{V} = 9.6 \, \text{V} \][/tex]
Thus, the P.d across the terminal is:
[tex]\[ \boxed{9.6 \, \text{V}} \][/tex]
The correct answer for the second problem is: B. [tex]\( 9.6 \, \text{V} \)[/tex]
Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.
