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Solve for r:

v = (4)/(3){r}^(3)
Solve for v2

a \: ave = (v2 - v1)/(t2 - t1)



1 Answer

3 votes

Answer:

Solving
v = (4)/(3)\pi r^3 for r gives us:
r=\sqrt[3]{(3v)/(4\pi)}

Solving
a_(ave) = (v_2-v_1)/(t_2-t_1) for v2 gives us:
v_2 = (t_2-t_1)a_(ave)+v_1

Explanation:

Solving an equation for a variable or constant means that we have to isolate the value on one side of the equation or write the whole equation in terms of that variable or constant.

Now,

Solving
v = (4)/(3)\pi r^3 for r


v = (4)/(3)\pi r^3

Multiplying whole equation by 3/4


(3)/(4).v = (3)/(4).(4)/(3) \pi r^3


(3)/(4)v = \pi r^3

Dividing by Pi on both sides


(3v)/(4\pi) = (\pi r^3)/(\pi)\\(3v)/(4\pi) = r^3

Taking cube root on both sides


\sqrt[3]{r^3} = \sqrt[3]{(3v)/(4\pi)} \\r = \sqrt[3]{(3v)/(4\pi)}

Now

Solving
a_(ave) = (v_2-v_1)/(t_2-t_1) for v2

Multiplying both sides by (t2-t1)


(t_2-t_1)a_(ave) = (v_2-v_1)/(t_2-t_1)(t_2-t_1)\\(t_2-t_1)a_(ave) = v_2-v_1

Adding v1 on both sides


(t_2-t_1)a_(ave)+v_1 = v_2-v_1+v_1\\(t_2-t_1)a_(ave)+v_1 = v_2

Hence,

Solving
v = (4)/(3)\pi r^3 for r gives us:
r=\sqrt[3]{(3v)/(4\pi)}

Solving
a_(ave) = (v_2-v_1)/(t_2-t_1) for v2 gives us:
v_2 = (t_2-t_1)a_(ave)+v_1

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