Answer:
This is because the rubbing releases negative charges, called electrons, which can build up on one object to produce a static charge. For example, when you shuffle your feet across a carpet, electrons can transfer onto you, building up a static charge on your skin.
Explanation:
This is because the rubbing releases negative charges
Frequency division multiplexing operates by dividing the signal into different frequencies
<u>Explanation:</u>
The technique that is used in the networking is the Frequency Division Multiplexing. using this technique, the existing bandwidths can be partitioned into different frequency bandwidths. These are not interrupting with each other. Each bandwidth can be used for carrying signals individually.
Using this technique many users can share a particular communication medium and they will not be interrupted with each other's communication.Hence this technique can also be termed as Frequency Division Multiple Access.
Explanation:
Given:
m = 1.673 × 10^-27 kg
Q = q = 1.602 × 10^-19 C
r = 0.75 nm
= 0.75 × 10^-9 m
A.
Energy, U = (kQq)/r
Ut = 1/2 mv^2 + 1/2 mv^2
1.673 × 10^-27 × v^2 = (8.99 × 10^9 × (1.602 × 10^-19)^2)/0.75 × 10^-9
v = 1.356 × 10^4 m/s
B.
F = (kQq)/r^2
F = m × a
1.673 × 10^-27 × a = ((8.99 × 10^9 × (1.602 × 10-19)^2)/(0.075 × 10^-9)^2
a = 2.45 × 10^17 m/s^2.
I think it is C. I hope I helped.
Data:
![f_{2} = 42 Hz](https://tex.z-dn.net/?f=f_%7B2%7D%20%3D%2042%20Hz)
n (Wave node)
V (Wave belly)
L (Wave length)
<span>The number of bells is equal to the number of the harmonic emitted by the string.
</span>
![f_{n} = \frac{nV}{2L}](https://tex.z-dn.net/?f=f_%7Bn%7D%20%3D%20%20%5Cfrac%7BnV%7D%7B2L%7D%20)
Wire 2 → 2º Harmonic → n = 2
![f_{n} = \frac{nV}{2L}](https://tex.z-dn.net/?f=f_%7Bn%7D%20%3D%20%5Cfrac%7BnV%7D%7B2L%7D%20)
![f_{2} = \frac{2V}{2L} ](https://tex.z-dn.net/?f=f_%7B2%7D%20%3D%20%5Cfrac%7B2V%7D%7B2L%7D%20%0A)
![2V = f_{2} *2L](https://tex.z-dn.net/?f=2V%20%3D%20%20f_%7B2%7D%20%2A2L)
![V = \frac{ f_{2}*2L }{2}](https://tex.z-dn.net/?f=V%20%3D%20%20%5Cfrac%7B%20f_%7B2%7D%2A2L%20%7D%7B2%7D%20)
![V = \frac{42*2L}{2}](https://tex.z-dn.net/?f=V%20%3D%20%20%5Cfrac%7B42%2A2L%7D%7B2%7D%20)
![V = \frac{84L}{2}](https://tex.z-dn.net/?f=V%20%3D%20%20%5Cfrac%7B84L%7D%7B2%7D%20)
![V = 42L](https://tex.z-dn.net/?f=V%20%3D%2042L)
Wire 1 → 1º Harmonic or Fundamental rope → n = 1
![f_{n} = \frac{nV}{2L}](https://tex.z-dn.net/?f=f_%7Bn%7D%20%3D%20%5Cfrac%7BnV%7D%7B2L%7D%20)
![f_{1} = \frac{1V}{2L}](https://tex.z-dn.net/?f=f_%7B1%7D%20%3D%20%5Cfrac%7B1V%7D%7B2L%7D%20)
![f_{1} = \frac{V}{2L}](https://tex.z-dn.net/?f=f_%7B1%7D%20%3D%20%20%5Cfrac%7BV%7D%7B2L%7D%20)
If, We have:
V = 42L
Soon:
![f_{1} = \frac{V}{2L}](https://tex.z-dn.net/?f=f_%7B1%7D%20%3D%20%5Cfrac%7BV%7D%7B2L%7D%20)
![f_{1} = \frac{42L}{2L}](https://tex.z-dn.net/?f=f_%7B1%7D%20%3D%20%5Cfrac%7B42L%7D%7B2L%7D%20)
![\boxed{f_{1} = 21 Hz}](https://tex.z-dn.net/?f=%5Cboxed%7Bf_%7B1%7D%20%3D%2021%20Hz%7D)
Answer:
<span>The fundamental frequency of the string:
</span>
21 Hz