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aleksley [76]
2 years ago
13

Create a ray diagram for eyeglasses that contain a diverging lens. Assume you are looking at a 2 cm tall object that is 4 cm fro

m the lens. The focal length is 3 cm.
Physics
1 answer:
k0ka [10]2 years ago
8 0

The ray diagram for the given object consists of 2 cm height of object, 4 cm object distance and 3 cm focal length.

<h3>Image formed by a diverging lens</h3>

Diverging lens is called a concave lens. The working of the lens is dependent on the refraction of the light rays as they pass through the lens.

Image formed by a diverging lens is always virtual, erect and diminished; smaller than the object and is located on the same side of the lens as the object.

The ray diagram for the given object is presented in the image in the diagram.

  • Object height = 2 cm
  • Focal length = 3 cm
  • Object distance = 4 cm

Learn more about diverging lens here: brainly.com/question/3140453

#SPJ1

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Consider the train car described in the previous part. Another experiment is conducted in it: A net force of 20N is applied to a
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3 years ago
A spring is 6.0cm long when it is not stretched, and 10cm long when a 7.0N force is applied. What force is needed to make it 20c
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Answer:

Approximately 25\; {\rm N} (assuming that this spring is ideal.)

Explanation:

The displacement of a spring is the new length of the spring relative to the original length.

For example:

  • When the 6.0\; {\rm cm}-spring in this question is stretched to 10\; {\rm cm}, the displacement is x = (10\; {\rm cm} - 6.0\; {\rm cm}).
  • Likewise, if this spring is stretched to 20\; {\rm cm}, the displacement would be (20\; {\rm cm} - 6\; {\rm cm}).

If this spring is ideal, the force on the spring would be proportional to the displacement of the spring. In other words, if a force of F_{\text{a}} displaces this spring by x_{\text{a}}, while a force of F_{\text{b}} displaces this spring by x_{\text{b}}, then:

\displaystyle \frac{F_{\text{a}}}{x_{\text{a}}} = \frac{F_{\text{b}}}{x_{\text{b}}}.

In this question, it is given that a force of F_{\text{a}} = 7.0 \; {\rm N} would stretch this spring by x_{\text{a}} = (10\; {\rm cm} - 6.0\; {\rm cm}). Thus, the force F_{\text{b}} required to stretch this spring by x_{\text{a}} = (20\; {\rm cm} - 6.0\; {\rm cm}) would satisfy:

\displaystyle \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}}= \frac{F_{\text{b}}}{20\; {\rm cm} - 6.0\; {\rm cm}}.

Rearrange and solve for F_{\text{b}}:

\begin{aligned} F_{\text{b}} &= \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}} \, (20\; {\rm cm} - 6.0\; {\rm cm}) \\ &\approx 25\; {\rm N}\end{aligned}.

7 0
2 years ago
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