Let's solve the given problem **step by step**: --- ## **Given:** - **Equation:** \[ MRR = \frac{KI}{T_m^{1.23}} \] - **Proportionality constant:** \( K = 40000 \, \frac{^\circ C^{1.23} \text{ mm}^3}{\text{amp} \cdot \text{min}} \) - **Melting point:** \( T_m = 310^\circ C \) - **Material Removal Rate:** \( MRR = .846 \, \text{mm}^3/\text{s} \) - **Find:** Current \( I \) (in amperes). --- ## **Step 1: Convert MRR to mm³/min** Since \( K \) is given in terms of minutes, convert MRR from \(\text{mm}^3/\text{s}\) to \(\text{mm}^3/\text{min}\): \[ MRR = .846 \, \frac{\text{mm}^3}{\text{s}} \times 60 \, \frac{\text{s}}{\text{min}} = 50.76\, \frac{\text{mm}^3}{\text{min}} \] --- ## **Step 2: Rearrange the Formula** Solve for current \( I \): \[ MRR = \frac{KI}{T_m^{1.23}} \implies I = \frac{MRR \cdot T_m^{1.23}}{K} \] --- ## **Step 3: Substitute the Values** - \( MRR = 50.76 \) - \( T_m = 310 \) - \( K = 40000 \) Plug in the values: \[ I = \frac{50.76 \times (310)^{1.23}}{40000} \] --- ## **Step 4: Calculate \( (310)^{1.23} \)** \[ \log_{10} 310 \approx 3.4914 \] \[ 1.23 \times 3.4914 \approx 4.2944 \] \[ 10^{4.2944} \approx 1.97 \times 10^4 \] But let's calculate it precisely: \[ 310^{1.23} = e^{1.23 \times \ln(310)} \] \[ \ln(310) \approx 8.043 \] \[ 1.23 \times 8.043 \approx 9.895 \] \[ e^{9.895} \approx 19780 \] --- ## **Step 5: Plug Back into the Formula** \[ I = \frac{50.76 \times 19780}{40000} \] \[ I = \frac{1004936.8}{40000} \] \[ I = 25.12 \, \text{A} \] --- ## **Final Answer** \[ \boxed{I = 25.1 \text{ amperes}} \] --- ### **Summary of Steps** 1. Convert MRR to mm³/min. 2. Rearranged the formula to solve for \( I \). 3. Plugged in the given values. 4. Calculated the exponent. 5. Evaluated the expression to find the current.