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The distance \( d \) that a spring is stretched by a hanging object varies directly as the weight \( w \) of the object. If a 27-kg object stretches a spring 53 cm, find the distance when the weight is 13 kg. Round your answer to the nearest whole centimeter.

A. 26 cm
B. 2 cm
C. 7 cm
D. 93 cm

Answer :

Final answer:

In this direct variation scenario, by finding the constant of variation with the initial given values, we can predict the distance a spring would stretch under a new weight. By applying these calculations, we find that a spring stretched by a 13-kg object would stretch approximately 26 cm.

Explanation:

The scenario described in your question is an example of direct variation, where the distance, d, that a spring is stretched varies directly with the weight, w. We can write this as a mathematical equation: d = kw, where k is the constant of variation.

We can find 'k' with the given information, where a 27-kg object stretches the spring 53 cm. This gives us: k = d/w = 53/27.

Once we have found 'k', we can find the distance the spring is stretched by a 13-kg object: d = kw = (53/27) * 13. Rounded to the nearest whole centimeter, we find the distance to be approximately 26 cm.A) 26 cm is the correct answer.

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