Answer :
The weight of the object on the planet with a free-fall acceleration of [tex]\(6.52 \, \text{m/s}^2\)[/tex] is approximately [tex]\(13.3 \, \text{N}\).[/tex]
To find the weight of the object on a planet with a free-fall acceleration of [tex]\(6.52 \, \text{m/s}^2\)[/tex], we can use the formula for weight:
[tex]\[ W = m \times g \][/tex]
Where:
- W is the weight of the object,
- m is the mass of the object, and
- g is the acceleration due to gravity.
Given that the weight of the object on Earth is [tex]\(4.50 \, \text{lb}\)[/tex], we first need to convert this to Newtons [tex](\(1 \, \text{lb} = 4.448 \, \text{N}\)):[/tex]
[tex]\[ W_{\text{Earth}} = 4.50 \, \text{lb} \times 4.448 \, \text{N/lb} \][/tex]
[tex]\[ W_{\text{Earth}} = 20.016 \, \text{N} \][/tex]
Now, we can use the weight formula to find the weight of the object on the planet with [tex]\(6.52 \, \text{m/s}^2\)[/tex] acceleration due to gravity:
[tex]\[ W_{\text{planet}} = m \times 6.52 \, \text{m/s}^2 \][/tex]
Given that weight is proportional to acceleration due to gravity, the weight on the new planet will be \(6.52/9.8\) times the weight on Earth. Thus,
[tex]\[ W_{\text{planet}} = \frac{6.52}{9.8} \times 20.016 \, \text{N} \][/tex]
[tex]\[ W_{\text{planet}} \approx 13.3 \, \text{N} \][/tex]
Therefore, the weight of the object on the planet with a free-fall acceleration of [tex]\(6.52 \, \text{m/s}^2\)[/tex] is approximately [tex]\(13.3 \, \text{N}\).[/tex]