The are elements on the periodic table
Just explain the day of how you were shopping and there you have it
For both NPN and PNP this is true:
The base is between the collector and the emitter.
The wavelength of the note is
![\lambda = 39.1 cm = 0.391 m](https://tex.z-dn.net/?f=%5Clambda%20%3D%2039.1%20cm%20%3D%200.391%20m)
. Since the speed of the wave is the speed of sound,
![c=344 m/s](https://tex.z-dn.net/?f=c%3D344%20m%2Fs)
, the frequency of the note is
![f= \frac{c}{\lambda}=879.8 Hz](https://tex.z-dn.net/?f=f%3D%20%5Cfrac%7Bc%7D%7B%5Clambda%7D%3D879.8%20Hz%20)
Then, we know that the frequency of a vibrating string is related to the tension T of the string and its length L by
![f= \frac{1}{2L} \sqrt{ \frac{T}{\mu} }](https://tex.z-dn.net/?f=f%3D%20%5Cfrac%7B1%7D%7B2L%7D%20%5Csqrt%7B%20%5Cfrac%7BT%7D%7B%5Cmu%7D%20%7D%20%20)
where
![\mu=0.550 g/m = 0.550 \cdot 10^{-3} kg/m](https://tex.z-dn.net/?f=%5Cmu%3D0.550%20g%2Fm%20%3D%200.550%20%5Ccdot%2010%5E%7B-3%7D%20kg%2Fm)
is the linear mass density of our string.
Using the value of the tension, T=160 N, and the frequency we just found, we can calculate the length of the string, L:
Answer:
The answer is 1.0 N
Explanation:
inclination of tray=12^{\circ}
gravitational Force=5 N
Now this gravitational force has two component i.e.
5\sin \theta is parallel to the tray =1.039 N
5\cos \theta is perpendicular to the tray =4.890 N