the formula for elastic potential energy, (p), of a spring is (p=\frac{1}{2}kx^{2}), where (k) is the spring…

the formula for elastic potential energy, (p), of a spring is (p=\frac{1}{2}kx^{2}), where (k) is the spring - constant and (x) is the length it stretches. a spring has a spring - constant of 0.51. which approximate formula can be used to determine the length it stretches, given (p)?\na. (x=sqrt{1.02p})\nb. (x = sqrt{2.47p})\nc. (x=sqrt{\frac{1.02}{p}})\nd. (x=sqrt{\frac{2.47}{p}})

the formula for elastic potential energy, (p), of a spring is (p=\frac{1}{2}kx^{2}), where (k) is the spring - constant and (x) is the length it stretches. a spring has a spring - constant of 0.51. which approximate formula can be used to determine the length it stretches, given (p)?\na. (x=sqrt{1.02p})\nb. (x = sqrt{2.47p})\nc. (x=sqrt{\frac{1.02}{p}})\nd. (x=sqrt{\frac{2.47}{p}})

Answer

Explanation:

Step1: Start with the formula

The formula for elastic - potential energy is $P=\frac{1}{2}kx^{2}$, where $k = 0.51$. We want to solve for $x$. First, multiply both sides of the equation by $2$ to get $2P=kx^{2}$.

Step2: Substitute the value of k

Since $k = 0.51$, the equation becomes $2P = 0.51x^{2}$.

Step3: Solve for x

Divide both sides by $0.51$: $x^{2}=\frac{2P}{0.51}$. Then take the square - root of both sides: $x=\sqrt{\frac{2P}{0.51}}$.

Answer:

D. $x=\sqrt{\frac{2P}{0.51}}$