Percent error is calculated as follows:
% = ( |15-15.6| / 15.6 ) * 100%
% = (0.6/15.6) * 100%
% = 0.0385 * 100%
% = 3.85%
Hope this helped!
Answer:
when work is done on the system or heat comes into the system
The spring is initially stretched, and the mass released from rest (v=0). The next time the speed becomes zero again is when the spring is fully compressed, and the mass is on the opposite side of the spring with respect to its equilibrium position, after a time t=0.100 s. This corresponds to half oscillation of the system. Therefore, the period of a full oscillation of the system is
![T=2 t = 2 \cdot 0.100 s = 0.200 s](https://tex.z-dn.net/?f=T%3D2%20t%20%3D%202%20%5Ccdot%200.100%20s%20%3D%200.200%20s)
Which means that the frequency is
![f= \frac{1}{T}= \frac{1}{0.200 s}=5 Hz](https://tex.z-dn.net/?f=f%3D%20%5Cfrac%7B1%7D%7BT%7D%3D%20%5Cfrac%7B1%7D%7B0.200%20s%7D%3D5%20Hz%20%20)
and the angular frequency is
![\omega=2 \pi f = 2 \pi (5 Hz)=31.4 rad/s](https://tex.z-dn.net/?f=%5Comega%3D2%20%5Cpi%20f%20%3D%202%20%5Cpi%20%285%20Hz%29%3D31.4%20rad%2Fs)
In a spring-mass system, the maximum velocity of the object is given by
![v_{max} = A \omega](https://tex.z-dn.net/?f=v_%7Bmax%7D%20%3D%20A%20%5Comega)
where A is the amplitude of the oscillation. In our problem, the amplitude of the motion corresponds to the initial displacement of the object (A=0.500 m), therefore the maximum velocity is
Answer:
yes This is correct Answer
Answer with Explanation:
The force of attraction between 2 charges of magnitude
separated by a distance 'r' is given by
where
is a constant known as permitivity of free space
![k=9\times 10^{9}Nm^2/C^2](https://tex.z-dn.net/?f=k%3D9%5Ctimes%2010%5E%7B9%7DNm%5E2%2FC%5E2)
Applying the given values in the above relation we get
![8=9\times 10^{9}\times \frac{130\times 180}{r^{2}}\\\\r^{2}=\frac{9\times 10^{9}\times 130\times 180}{8}\\\\r^{2}=26325\times 10^{9}\\\\\therefore r=\sqrt{26325\times 10^{9}}=5.131\times 10^{6}meters](https://tex.z-dn.net/?f=8%3D9%5Ctimes%2010%5E%7B9%7D%5Ctimes%20%5Cfrac%7B130%5Ctimes%20180%7D%7Br%5E%7B2%7D%7D%5C%5C%5C%5Cr%5E%7B2%7D%3D%5Cfrac%7B9%5Ctimes%2010%5E%7B9%7D%5Ctimes%20130%5Ctimes%20180%7D%7B8%7D%5C%5C%5C%5Cr%5E%7B2%7D%3D26325%5Ctimes%2010%5E%7B9%7D%5C%5C%5C%5C%5Ctherefore%20r%3D%5Csqrt%7B26325%5Ctimes%2010%5E%7B9%7D%7D%3D5.131%5Ctimes%2010%5E%7B6%7Dmeters)