ron weighs a bottle with a mass of 0.5 kilograms on a spring scale. the spring stretches by 1 centimeter. he…

Explanation:

Step1: Establish the proportional - relationship

According to Hooke's law, the force exerted on the spring is proportional to the extension of the spring, and the force (F = mg). For the first bottle, (F_1=m_1g) and the extension (x_1). For the second bottle, (F_2 = m_2g) and the extension (x_2). Since (\frac{F_1}{x_1}=\frac{F_2}{x_2}) (because the spring constant (k) is the same for the same spring), and (F = mg), we have (\frac{m_1g}{x_1}=\frac{m_2g}{x_2}), which simplifies to (\frac{m_1}{x_1}=\frac{m_2}{x_2}).

Step2: Substitute the known values

We know that (m_1 = 0.5\space kg), (x_1=1\space cm), and (x_2 = 2\space cm). Substituting these values into (\frac{m_1}{x_1}=\frac{m_2}{x_2}), we get (m_2=\frac{m_1x_2}{x_1}).

Step3: Calculate the mass of the second bottle

(m_2=\frac{0.5\times2}{1}=1\space kg).

Answer:

1 kg