the length of a spring varies directly with the mass of an object that is attached to it. when a 30 - gram…

the length of a spring varies directly with the mass of an object that is attached to it. when a 30 - gram object is attached, the spring stretches 9 centimeters. which equation relates the mass of the object, m, and the length of the spring, s?\n$ s=\frac{3}{10}m$\n$ s = \frac{10}{3}m$\n$ m=\frac{3}{10}s$\n$ m=\frac{1}{30}s$

the length of a spring varies directly with the mass of an object that is attached to it. when a 30 - gram object is attached, the spring stretches 9 centimeters. which equation relates the mass of the object, m, and the length of the spring, s?\n$ s=\frac{3}{10}m$\n$ s = \frac{10}{3}m$\n$ m=\frac{3}{10}s$\n$ m=\frac{1}{30}s$

Answer

Answer:

A. $s=\frac{3}{10}m$

Explanation:

Step1: Recall direct - variation formula

The direct - variation formula is $s = km$, where $s$ is the length of the spring, $m$ is the mass of the object, and $k$ is the constant of variation.

Step2: Find the value of $k$

We know that when $m = 30$ grams and $s=9$ centimeters. Substitute these values into $s = km$: $9=k\times30$. Solve for $k$ by dividing both sides by 30: $k=\frac{9}{30}=\frac{3}{10}$.

Step3: Write the equation

Substitute $k = \frac{3}{10}$ back into the direct - variation formula $s = km$. We get $s=\frac{3}{10}m$.