Respuesta :
Answer:
The answer is "[tex]1.75 \cdot (2.60 \hat{i} + 3.5 \hat{j}) \times 10^{11} \ \ \frac{m}{s^2} \\[/tex] ".
Explanation:
Formula of acceleration =
[tex]\frac{F_e}{m_e} =\frac{-e(\underset{E}{\rightarrow} + \underset{V}{\rightarrow} \times \underset{B}{\rightarrow})}{m_e}[/tex]
values:
[tex]\underset{E}{\rightarrow} = (2.60 \hat{i} + 5.90 \hat{j}) \frac{V}{m} \\\\\underset{B}{\rightarrow} = 0.400 k \ T \\\\\underset{V}{\rightarrow} = 6.0 \hat {i} \ \ \frac{m}{s} \\\\\ apply \ value \ in \ above \ formula: \\\\ \frac{F_e}{m_e} =\frac{e}{m_e} \cdot (2.60 \hat{i} + 5.90 \hat{j}+6.0 \hat {i} \times 0.4\hat {k} ) \\\\ \frac{F_e}{m_e} =\frac{e}{m_e} \cdot (2.60 \hat{i} + 5.90 \hat{j}- 2.4 \hat {j}) \\\\\therefore \frac{e}{m_e} = \frac{1.6 \times 10^{-19}}{9.1 \times 10^{-21}} \\\\[/tex]
[tex]\frac{F_e}{m_e} = \frac{1.6 \times 10^{-19}}{9.1 \times 10^{-21}} \cdot (2.60 \hat{i} + 5.90 \hat{j}- 2.4 \hat {j}) \\\\\frac{F_e}{m_e} = \frac{1.6 \times 10^{-19}}{9.1 \times 10^{-21}} \cdot (2.60 \hat{i} + 3.5 \hat{j}) \\\\\frac{F_e}{m_e} = 1.75 \cdot (2.60 \hat{i} + 3.5 \hat{j}) \times 10^{11} \ \ \frac{m}{s^2} \\\\[/tex]
