False, applied force is when a person or an object pushes on another object
If the object is in equilibrium that means that the sum of the forces on it is zero and the net force is zero. If none of the forces changes then the object continues in constant uniform motion. That means constant speed in a straight line.
Answer: C) Increase the amplitude of the wavelenghth to increase the intensity.
Explanation:
Answer:
(a) I=0.01 kg.m²
(b) I=0.03 kg.m²
Explanation:
Given data
Mass of disk M=2.0 kg
Diameter of disk d=20 cm=0.20 m
To Find
(a) Moment of inertia through the center of disk
(b) Moment of inertia through the edge of disk
Solution
For (a) Moment of inertia through the center of disk
Using the equation of moment of Inertia
![I=\frac{1}{2}MR^{2}\\ I=\frac{1}{2}(2.0kg)(0.20m/2)^{2}\\ I=0.01 kg m^{2}](https://tex.z-dn.net/?f=I%3D%5Cfrac%7B1%7D%7B2%7DMR%5E%7B2%7D%5C%5C%20%20I%3D%5Cfrac%7B1%7D%7B2%7D%282.0kg%29%280.20m%2F2%29%5E%7B2%7D%5C%5C%20%20I%3D0.01%20kg%20m%5E%7B2%7D)
For (b) Moment of inertia through the edge of disk
We can apply parallel axis theorem for calculating moment of inertia
Explanation:
Formula to determine the critical crack is as follows.
![K_{IC} = \gamma \sigma_{f} \sqrt{\pi \times a}](https://tex.z-dn.net/?f=K_%7BIC%7D%20%3D%20%5Cgamma%20%5Csigma_%7Bf%7D%20%5Csqrt%7B%5Cpi%20%5Ctimes%20a%7D)
= 1,
= 24.1
[/tex]\sigma_{y}[/tex] = 570
and, ![\sigma_{f} = 570 \times \frac{3}{4}](https://tex.z-dn.net/?f=%5Csigma_%7Bf%7D%20%3D%20570%20%5Ctimes%20%5Cfrac%7B3%7D%7B4%7D)
= 427.5
Hence, we will calculate the critical crack length as follows.
a = ![\frac{1}{\pi} \times (\frac{K_{IC}}{\sigma_{f}})^{2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5Cpi%7D%20%5Ctimes%20%28%5Cfrac%7BK_%7BIC%7D%7D%7B%5Csigma_%7Bf%7D%7D%29%5E%7B2%7D)
= ![\frac{1}{3.14} \times (\frac{24.1}{427.5})^{2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B3.14%7D%20%5Ctimes%20%28%5Cfrac%7B24.1%7D%7B427.5%7D%29%5E%7B2%7D)
= ![10.13 \times 10^{-4}](https://tex.z-dn.net/?f=10.13%20%5Ctimes%2010%5E%7B-4%7D)
Therefore, largest size is as follows.
Largest size = 2a
= ![2 \times 10.13 \times 10^{-4}](https://tex.z-dn.net/?f=2%20%5Ctimes%2010.13%20%5Ctimes%2010%5E%7B-4%7D)
= ![20.26 \times 10^{-4}](https://tex.z-dn.net/?f=20.26%20%5Ctimes%2010%5E%7B-4%7D)
Thus, we can conclude that the critical crack length for a through crack contained within the given plate is
.