**Question:**The block is at rest as shown. What is the period of the oscillation if the block is pulled down by 10 cm, in seconds? Use \( g = 10 \, \text{m/s}^2 \).*Your answer needs to have 2 significant figures, including the negative sign in your answer if needed. Do not include the positive sign if the answer is positive. No unit is needed in your answer; it is already given in the question statement.***Diagram Explanation:**The diagram shows a vertical spring with a block attached to its lower end. The following details are included in the diagram:- A vertical ruler is positioned adjacent to the spring, marked from 0 to 100 cm.- The scale indicates various measurements on the ruler at 10 cm increments.- The block, labeled as "100 g," hangs from the spring at a point slightly below the 90 cm mark.- The spring is slightly stretched due to the weight of the block.This setup illustrates a typical spring-mass system used to study oscillations and the effect of gravity on a suspended mass. **Problem Statement:**A simple pendulum of mass \( m = 2.00 \, \text{kg} \) and length \( L = 0.82 \, \text{m} \) on planet X, where the value of \( g \) is unknown, oscillates with a period \( T = 1.70 \, \text{s} \). What is the period if the length is doubled?**Options:**- \( 2.4 \, \text{s} \)- \( 0.85 \, \text{s} \)- \( 1.2 \, \text{s} \)- \( 1.7 \, \text{s} \)- \( 3.4 \, \text{s} \)This question is designed to test your understanding of the relationship between the length of a pendulum and its period of oscillation. The period of a simple pendulum is given by the formula:\[T = 2\pi \sqrt{\frac{L}{g}}\]Where:- \( T \) is the period,- \( L \) is the length of the pendulum,- \( g \) is the acceleration due to gravity.If the length \( L \) is doubled, we need to determine the new period.

Answered: Image /qna-images/answer/c51d6f85-69ac-4a16-9915-2a88d8e96760.jpg

The block is at rest as shown. What is the peiod of the oscillation if the block is pulled down by 10 cm, in seconds? Use g = 10 m/s². Your answer needs to have 2 significant figures, including the negative sign in your answer if needed. Do not include the positive sign if the answer is positive. No unit is needed in your answer, it is already given in the question statement. 100 g ā 8

The block is at rest as shown. What is the peiod of the oscillation if the block is pulled down by 10 cm, in seconds? Use g = 10 m/s². Your answer needs to have 2 significant figures, including the negative sign in your answer if needed. Do not include the positive sign if the answer is positive. No unit is needed in your answer, it is already given in the question statement. 100 g ā 8

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Estimate of calculation

Question

**Problem Statement:**

A simple pendulum of mass \( m = 2.00 \, \text{kg} \) and length \( L = 0.82 \, \text{m} \) on planet X, where the value of \( g \) is unknown, oscillates with a period \( T = 1.70 \, \text{s} \). What is the period if the length is doubled?

**Options:**

- \( 2.4 \, \text{s} \)
- \( 0.85 \, \text{s} \)
- \( 1.2 \, \text{s} \)
- \( 1.7 \, \text{s} \)
- \( 3.4 \, \text{s} \)

This question is designed to test your understanding of the relationship between the length of a pendulum and its period of oscillation. The period of a simple pendulum is given by the formula:

\[
T = 2\pi \sqrt{\frac{L}{g}}
\]

Where:
- \( T \) is the period,
- \( L \) is the length of the pendulum,
- \( g \) is the acceleration due to gravity.

If the length \( L \) is doubled, we need to determine the new period.

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