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Solve for k J=k24
(a)k=±2J√J
(b)k=±4J√
(c)±4J
(d)k=±2J√

2 Answers

0 votes

Here are a bunch of CORRECT answers. Your answer is in the second pic. I got number 3 wrong, but it still showed the correct answer.

Solve for k J=k24 (a)k=±2J√J (b)k=±4J√ (c)±4J (d)k=±2J√-example-1
Solve for k J=k24 (a)k=±2J√J (b)k=±4J√ (c)±4J (d)k=±2J√-example-2
Solve for k J=k24 (a)k=±2J√J (b)k=±4J√ (c)±4J (d)k=±2J√-example-3
Solve for k J=k24 (a)k=±2J√J (b)k=±4J√ (c)±4J (d)k=±2J√-example-4
Solve for k J=k24 (a)k=±2J√J (b)k=±4J√ (c)±4J (d)k=±2J√-example-5
User Sluukkonen
by
8.1k points
2 votes

Answer:


k=2√(J)

Explanation:

If the equation is
J = (k^2)/(4), begin by taking the square root of each side.


√(J) = \sqrt{(k^2)/(4)} \\√(J)=\frac {k}{2}

Now multiply each side by 2.


√(J)=\frac {k}{2}\\2√(J) = k

The solution is
k=2√(J) or option d.

User Mat Kelcey
by
8.7k points