contestada

An uncharged molecule of DNA (deoxyribonucleic acid) is 2.14 µm long. The ends of the molecule become singly ionized so that there is a charge of −1.6 × 10−19 C on one end and +1.6 × 10−19 C on the other. The helical molecule acts like a spring and compresses 1.4% upon becoming charged. Find the effective spring constant of the molecule. The value of Coulomb’s constant is 8.98755 × 109 N · m2 /C 2 and the acceleration due to gravity is 9.8 m/s 2 . Answer in units of N/m.

Respuesta :

Answer:

K = 1.72 *10^{-11} N/m

Explanation:

given data:

L= 2.14*10^{-6} m

[tex]\Delta = 1.4%* of L[/tex]

           [tex]= \frac{1.4}{100} *2.14*10^{-6}[/tex]

           = 2.99*10^{-8} m

[tex]L' = L - \Delta L[/tex]

[tex]L ' = 2.14*10^{-6} -2.99*10^{-8}[/tex]

[tex]L = 2.11*10^{-6} m[/tex]

electrostatic force = spring force

[tex]\frac{ kq^2}{L'^2} = K \Delta L[/tex]

putting all equation to get spring constant K value

[tex]\frac{8.98*10^9 *1.6*10^{-19}}{2.11*10^{-6}} = K 2.99*10^{-8}[/tex]

K = 1.72 *10^{-11} N/m

Otras preguntas