Respuesta :
To calculate the tension in the spring when the object is immersed in the liquid, we need to consider the buoyant force acting on the object.
First, let's find the volume of the object immersed in the liquid.
Given:
- Density of the object (\( \rho_{\text{object}} \)) = 20 g/cm³
- Weight of the object (\( w \)) = 200 g
- Density of the liquid (\( \rho_{\text{liquid}} \)) = \( \frac{7}{3} \times \rho_{\text{object}} = \frac{7}{3} \times 20 \) g/cm³
The volume of the object immersed in the liquid (\( V_{\text{immersed}} \)) can be calculated using the formula for density:
\[ \rho_{\text{object}} = \frac{m_{\text{object}}}{V_{\text{object}}} \]
where \( m_{\text{object}} \) is the mass of the object and \( V_{\text{object}} \) is the volume of the object.
We can rearrange the formula to solve for \( V_{\text{object}} \):
\[ V_{\text{object}} = \frac{m_{\text{object}}}{\rho_{\text{object}}} \]
Given that the weight of the object (\( w \)) is 200 g, we can find the mass of the object (\( m_{\text{object}} \)) using the formula \( w = m_{\text{object}} \times g \), where \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).
\[ m_{\text{object}} = \frac{w}{g} \]
Now, we can calculate the volume of the object (\( V_{\text{object}} \)):
\[ V_{\text{object}} = \frac{m_{\text{object}}}{\rho_{\text{object}}} \]
Next, we need to calculate the volume of the object immersed in the liquid (\( V_{\text{immersed}} \)):
\[ V_{\text{immersed}} = \frac{\rho_{\text{object}}}{\rho_{\text{liquid}}} \times V_{\text{object}} \]
Once we have the volume of the object immersed in the liquid, we can calculate the buoyant force (\( F_{\text{buoyant}} \)) exerted on the object using the formula:
\[ F_{\text{buoyant}} = \rho_{\text{liquid}} \times g \times V_{\text{immersed}} \]
Finally, the tension in the spring (\( T \)) is equal to the weight of the object minus the buoyant force:
\[ T = w - F_{\text{buoyant}} \]
Let's calculate these values. First, let's calculate the mass of the object (\( m_{\text{object}} \)):
\[ m_{\text{object}} = \frac{w}{g} = \frac{200 \, \text{g}}{9.81 \, \text{m/s}^2} \]
\[ m_{\text{object}} \approx 20.37 \, \text{kg} \]
Next, let's calculate the volume of the object (\( V_{\text{object}} \)):
\[ V_{\text{object}} = \frac{m_{\text{object}}}{\rho_{\text{object}}} = \frac{20.37 \, \text{kg}}{20 \, \text{g/cm}^3} \]
\[ V_{\text{object}} \approx 1.0185 \, \text{cm}^3 \]
Now, let's calculate the volume of the object immersed in the liquid (\( V_{\text{immersed}} \)):
\[ V_{\text{immersed}} = \frac{\rho_{\text{object}}}{\rho_{\text{liquid}}} \times V_{\text{object}} = \frac{20 \, \text{g/cm}^3}{\frac{7}{3} \times 20 \, \text{g/cm}^3} \times 1.0185 \, \text{cm}^3 \]
\[ V_{\text{immersed}} = \frac{3}{7} \times 1.0185 \, \text{cm}^3 \]
\[ V_{\text{immersed}} \approx 0.4376 \, \text{cm}^3 \]
Now, let's calculate the buoyant force (\( F_{\text{buoyant}} \)):
\[ F_{\text{buoyant}} = \rho_{\text{liquid}} \times g \times V_{\text{immersed}} = \left( \frac{7}{3} \times 20 \, \text{g/cm}^3 \right) \times 9.81 \, \text{m/s}^2 \times 0.4376 \, \text{cm}^3 \]
\[ F_{\text{buoyant}} \approx 137.33 \, \text{N} \]
Finally, let's calculate the tension in the spring (\( T \)):
\[ T = w - F_{\text{buoyant}} = 200 \, \text{N} - 137.33 \, \text{N} \]
\[ T \approx 62.67 \, \text{N} \]
So, the tension in the spring is approximately \( 62.67 \, \text{N} \).
First, let's find the volume of the object immersed in the liquid.
Given:
- Density of the object (\( \rho_{\text{object}} \)) = 20 g/cm³
- Weight of the object (\( w \)) = 200 g
- Density of the liquid (\( \rho_{\text{liquid}} \)) = \( \frac{7}{3} \times \rho_{\text{object}} = \frac{7}{3} \times 20 \) g/cm³
The volume of the object immersed in the liquid (\( V_{\text{immersed}} \)) can be calculated using the formula for density:
\[ \rho_{\text{object}} = \frac{m_{\text{object}}}{V_{\text{object}}} \]
where \( m_{\text{object}} \) is the mass of the object and \( V_{\text{object}} \) is the volume of the object.
We can rearrange the formula to solve for \( V_{\text{object}} \):
\[ V_{\text{object}} = \frac{m_{\text{object}}}{\rho_{\text{object}}} \]
Given that the weight of the object (\( w \)) is 200 g, we can find the mass of the object (\( m_{\text{object}} \)) using the formula \( w = m_{\text{object}} \times g \), where \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).
\[ m_{\text{object}} = \frac{w}{g} \]
Now, we can calculate the volume of the object (\( V_{\text{object}} \)):
\[ V_{\text{object}} = \frac{m_{\text{object}}}{\rho_{\text{object}}} \]
Next, we need to calculate the volume of the object immersed in the liquid (\( V_{\text{immersed}} \)):
\[ V_{\text{immersed}} = \frac{\rho_{\text{object}}}{\rho_{\text{liquid}}} \times V_{\text{object}} \]
Once we have the volume of the object immersed in the liquid, we can calculate the buoyant force (\( F_{\text{buoyant}} \)) exerted on the object using the formula:
\[ F_{\text{buoyant}} = \rho_{\text{liquid}} \times g \times V_{\text{immersed}} \]
Finally, the tension in the spring (\( T \)) is equal to the weight of the object minus the buoyant force:
\[ T = w - F_{\text{buoyant}} \]
Let's calculate these values. First, let's calculate the mass of the object (\( m_{\text{object}} \)):
\[ m_{\text{object}} = \frac{w}{g} = \frac{200 \, \text{g}}{9.81 \, \text{m/s}^2} \]
\[ m_{\text{object}} \approx 20.37 \, \text{kg} \]
Next, let's calculate the volume of the object (\( V_{\text{object}} \)):
\[ V_{\text{object}} = \frac{m_{\text{object}}}{\rho_{\text{object}}} = \frac{20.37 \, \text{kg}}{20 \, \text{g/cm}^3} \]
\[ V_{\text{object}} \approx 1.0185 \, \text{cm}^3 \]
Now, let's calculate the volume of the object immersed in the liquid (\( V_{\text{immersed}} \)):
\[ V_{\text{immersed}} = \frac{\rho_{\text{object}}}{\rho_{\text{liquid}}} \times V_{\text{object}} = \frac{20 \, \text{g/cm}^3}{\frac{7}{3} \times 20 \, \text{g/cm}^3} \times 1.0185 \, \text{cm}^3 \]
\[ V_{\text{immersed}} = \frac{3}{7} \times 1.0185 \, \text{cm}^3 \]
\[ V_{\text{immersed}} \approx 0.4376 \, \text{cm}^3 \]
Now, let's calculate the buoyant force (\( F_{\text{buoyant}} \)):
\[ F_{\text{buoyant}} = \rho_{\text{liquid}} \times g \times V_{\text{immersed}} = \left( \frac{7}{3} \times 20 \, \text{g/cm}^3 \right) \times 9.81 \, \text{m/s}^2 \times 0.4376 \, \text{cm}^3 \]
\[ F_{\text{buoyant}} \approx 137.33 \, \text{N} \]
Finally, let's calculate the tension in the spring (\( T \)):
\[ T = w - F_{\text{buoyant}} = 200 \, \text{N} - 137.33 \, \text{N} \]
\[ T \approx 62.67 \, \text{N} \]
So, the tension in the spring is approximately \( 62.67 \, \text{N} \).
