1) Dane [ gamma_(4)=6 (mathrm(k))/(mathrm(c)) gamma_(2)=2 (mathrm(k))/(mathrm(c)) m_(4)=40 mathrm(k) (m_(2)=20 mathrm(~m))/(x_(x)-?) ]

It seems like there might be a typo in the last line of the provided data. Let's correct it and proceed with the calculations.<br /><br />Given:<br />\[<br />\begin{array}{l}<br />\gamma_{4}=6 \frac{\mathrm{k}}{\mathrm{c}} \\<br />\gamma_{2}=2 \frac{\mathrm{k}}{\mathrm{c}} \\<br />m_{4}=40 \mathrm{k} \\<br />m_{2}=20 \mathrm{k}<br />\end{array}<br />\]<br /><br />To find \( x_{x} \), we need to understand the relationship between these variables. Assuming \( x_{x} \) is the acceleration, we can use Newton's second law:<br /><br />\[ F = ma \]<br /><br />For object 4:<br />\[ F_4 = m_4 \cdot \gamma_4 = 40 \mathrm{k} \cdot 6 \frac{\mathrm{k}}{\mathrm{c}} = 240 \mathrm{k}^2/\mathrm{c} \]<br /><br />For object 2:<br />\[ F_2 = m_2 \cdot \gamma_2 = 20 \mathrm{k} \cdot 2 \frac{\mathrm{k}}{\mathrm{c}} = 40 \mathrm{k}^2/\mathrm{c} \]<br /><br />Since the forces are equal (assuming no external forces acting on the system):<br />\[ F_4 = F_2 \]<br /><br />Thus, the acceleration \( x_{x} \) is the same for both objects:<br />\[ x_{x} = \gamma_4 = 6 \frac{\mathrm{k}}{\mathrm{c}} \]<br /><br />So, the value of \( x_{x} \) is \( 6 \frac{\mathrm{k}}{\mathrm{c}} \).