Just learned this in chemistry. There are no subscripts, therefor each element only has 1 atom. Na has 1, O has 1 and H has one. There are 3 atoms.
Answer-3
Answer:
Part A:
![E=127500N/C\\E=1.275*10^{5} N/C](https://tex.z-dn.net/?f=E%3D127500N%2FC%5C%5CE%3D1.275%2A10%5E%7B5%7D%20N%2FC)
Part B:
Option B (Towards the South)
Explanation:
Part A:
Magnitude if electric field E:
E=Force/charge
Force=2.04×10−14 N
Charge=1.6×10−19 C
![E=\frac{2.04*10^{-14}}{1.6*10^{-19}} \\E=127500N/C\\E=1.275*10^{5} N/C](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B2.04%2A10%5E%7B-14%7D%7D%7B1.6%2A10%5E%7B-19%7D%7D%20%5C%5CE%3D127500N%2FC%5C%5CE%3D1.275%2A10%5E%7B5%7D%20N%2FC)
Part B:
Option B (Towards the South)
As electron is experiencing the force towards south,it means the direction of the electric field is towards the south because direction of field lines is from positive to negative, so proton is moving towards south it means negative charge is in south to which proton is attracted. So electric field is towards South.
Answer:
Potential energy is ![U=mgh](https://tex.z-dn.net/?f=U%3Dmgh)
Explanation:
The potential energy depends on the mass, the acceleration of gravity g and the height at which the object or person is.
Potential energy ![U=mgh](https://tex.z-dn.net/?f=U%3Dmgh)
In this case we would need to know the exact mass of the hiker in order to calculate the potential energy.
But we know the values of g and h
![g=9.81m/s^2](https://tex.z-dn.net/?f=g%3D9.81m%2Fs%5E2)
![h=60m](https://tex.z-dn.net/?f=h%3D60m)
So, the potential energy
![U=m(9.81m/s^2)(60m)\\\\U=588.6*m](https://tex.z-dn.net/?f=U%3Dm%289.81m%2Fs%5E2%29%2860m%29%5C%5C%5C%5CU%3D588.6%2Am)
m is the mass of the hiker, wich is not in the description of the problem.
Answer:
anyone know this or will i have to get my brother
Answer:
Frequency of oscillation, f = 4 Hz
time period, T = 0.25 s
Angular frequency, ![\omega = 25.13 rad/s](https://tex.z-dn.net/?f=%5Comega%20%3D%2025.13%20rad%2Fs)
Given:
Time taken to make one oscillation, T = 0.25 s
Solution:
Frequency, f of oscillation is given as the reciprocal of time taken for one oscillation and is given by:
f = ![\frac{1}{T}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7BT%7D)
f = ![\frac{1}{0.25}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B0.25%7D)
Frequency of oscillation, f = 4 Hz
The period of oscillation can be defined as the time taken by the suspended mass for completion of one oscillation.
Therefore, time period, T = 0.25 s
Angular frequency of oscillation is given by:
![\omega = 2\pi \times f](https://tex.z-dn.net/?f=%5Comega%20%3D%202%5Cpi%20%5Ctimes%20f)
![\omega = 2\pi \times 4](https://tex.z-dn.net/?f=%5Comega%20%3D%202%5Cpi%20%5Ctimes%204)
![\omega = 25.13 rad/s](https://tex.z-dn.net/?f=%5Comega%20%3D%2025.13%20rad%2Fs)