Using the data of Table 19.1, determine the length in inches of a simple pendulum that oscillates with a period of 2 s and an amplitude of 90°.Reference to Table 19.1: Table 19.1 Correction Factor for the Period of a SimplePendulumӨт0°10°20°30°60°90°120°150°180°K1.5711.5741.5831.5981.6861.8542.1572.7682K/T1.0001.0021.008 1.0171.0731.1801.3731.762

Write the formula for the time period of the pendulum for large oscillations. Here, is the time…

Table 19.1 Correction Factor for the Period of a Simple Pendulum Өт 0° 10° 20° 30° 60° 90° 120° 150° 180° K 1.571 1.574 1.583 1.598 1.686 1.854 2.157 2.768 2K/T 1.000 1.002 1.008 1.017 1.073 1.180 1.373 1.762

Table 19.1 Correction Factor for the Period of a Simple Pendulum Өт 0° 10° 20° 30° 60° 90° 120° 150° 180° K 1.571 1.574 1.583 1.598 1.686 1.854 2.157 2.768 2K/T 1.000 1.002 1.008 1.017 1.073 1.180 1.373 1.762

Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)

8th Edition

ISBN:9781305387102

Author:Kreith, Frank; Manglik, Raj M.

Publisher:Kreith, Frank; Manglik, Raj M.

Chapter5: Analysis Of Convection Heat Transfer

Section: Chapter Questions

Problem 5.2P: 5.2 Evaluate the Prandtl number from the following data: , .

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Concept explainers

Strain Gauges

All the components used in civil engineering and mechanical engineering applications undergo deformations due to induced internal stresses under the application of external loads. Under repeated application of loads, there is a high possibility of failure of the components. Hence, periodic monitoring of the components and determination of stresses are very important.

Question

Using the data of Table 19.1, determine the length in inches of a simple pendulum that oscillates with a period of 2 s and an amplitude of 90°.

Reference to Table 19.1:

Table 19.1 Correction Factor for the Period of a Simple
Pendulum
Өт
0°
10°
20°
30°
60°
90°
120°
150°
180°
K
1.571
1.574
1.583
1.598
1.686
1.854
2.157
2.768
2K/T
1.000
1.002
1.008 1.017
1.073
1.180
1.373
1.762

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