Westonci.ca is the best place to get answers to your questions, provided by a community of experienced and knowledgeable experts. Discover the answers you need from a community of experts ready to help you with their knowledge and experience in various fields. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
Sagot :
The weight of the object hung from the spring is 6.27 N.
Hooke's law;
- Hooke's law states that the load applied to an elastic material is directly proportional to the extension of the material provided elastic limit is not exceeded.
F = Kx
[tex]k = \frac{F}{x} \\\\k = F (\frac{1}{x} )\\\\k = \frac{1}{slope}[/tex]
From the graph, we can use the following points to determine the slope of the graph;
(2, 0.8) and (4, 1.5)
[tex]k = \frac{1.5 - 0.8}{4-2} \\\\k = 0.35 \\\\\frac{1}{k} = \frac{1}{0.35} = 2.85 \ N/cm[/tex]
The extension of the spring is calculated as follows;
x = 19.2 cm - 17.0 cm = 2.2 cm
The weight of the object hung from the spring is calculated as follows;
F = kx
F = 2.85 x 2.2
F = 6.27 N
Learn more about Hooke's law here: https://brainly.com/question/2648431

Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.