Physics: What ratio increases when a cell divides into two smaller cells?

What ratio increases when a cell divides into two

Ответ:

summerjoiner

30.07.2021

The initial speed of the father is 1.9 m/s and the initial speed of the son is 4.65 m/s.

The given parameters;

kinetic of the son = K.Ekinetic energy of the father = $\frac{K.E}{3}$ mass of the father = mmass of the son = $\frac{m}{2}$ father's final speed = v + 1.4let the initial speed of father = v

The kinetic energy of the son is written as follows;

$K.E_s = \frac{1}{2} m v^2$

The kinetic energy of the father is written as follows;

$K.E_f =\frac{1}{3} K.E_s$

When the fathers speed increases by 1.4 m/s;

$\frac{1}{2} mv^2 = \frac{1}{3} [\frac{1}{2} m(v + 1.4)^2]\\\\ v^2 = \frac{1}{3} (v^2 + 2.8v + 1.96)\\\\ v^2 = \frac{v^2}{3} + 0.93v + 0.653\\\\\frac{2}{3} v^2 - 0.93v - 0.653 = 0\\\\solve \ quadratic \ equation \ using \ formula \ method;\\\\a = \frac{2}{3} , \ b = -0.93 , \ \ c = -0.653\\\\v = \frac{-b \ \ +/- \sqrt{b^2 - 4ac} }{2a} \\\\v = \frac{-(-0.93) \ \ +/- \sqrt{(-0.93)^2 - 4(\frac{2}{3} \times -0.653)} }{2(\frac{2}{3})}\\\\v = 1.9 \ m/s$

The velocity of the son is calculated as follows;

$K.E_f = \frac{1}{3} K.E_s\\\\\frac{1}{2} mv_f^2 = \frac{1}{3} [\frac{1}{2} . \frac{m}{2} v_s^2]\\\\v_f^2 = \frac{1}{6} v_s^2\\\\v_s^2 = 6v_f^2\\\\v_s^2 = 6(1.9)^2\\\\v_s^2 = 21.66 \\\\v_s = \sqrt{21.66} \\\\v_s = 4.65\ m/s$

Thus, the initial speed of the father is 1.9 m/s and the initial speed of the son is 4.65 m/s.

Learn more here:link

0,0(0 оценок)

What ratio increases when a cell divides into two smaller cells?

Ответ:

Ответ:

Ask your question to the AI advisor

More tips

Answers on questions: Physics

Ask an AI advisor a question