Answer:
The answer is "[tex]4,500 - 225 \ V[/tex]".
Explanation:
Using formula for calculating the Voltage:
[tex]M_1=12.5\\\\M_2=250\\\\V_1=4,500 \\\\\bold{\text{Formula: }}\\\\\to \bold{\frac{m_1}{m_2}=\frac{V_2}{V_1}}\\\\\to \frac{12.5}{250}=\frac{V_2}{4,500}\\\\\to 0.05=\frac{v_2}{4,500}\\\\\to 0.05\times 4,500= V_2\\\\\to V_2=225\\\\[/tex]
Hence the range of accelerating in voltage is [tex]4,500 - 225 \ V[/tex]