Answer:
a) E=228391.8 N/C
b) E=-59345.91N/C
Explanation:
You can use Gauss law to find the net electric field produced by both line of charges.

Where E1 and E2 are the electric field generated at a distance of r1 and r2 respectively from the line of charges.
The net electric field at point r will be:

a) for y=0.200m, r1=0.200m and r2=0.200m:
![E=\frac{1}{2\pi(8.85*10^{-12}C^2/Nm^2)}[\frac{4.80*10^{-6}C}{0.200m}-\frac{2.26*10^{-6}C}{0.200m}}]=228391.8N/C](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B1%7D%7B2%5Cpi%288.85%2A10%5E%7B-12%7DC%5E2%2FNm%5E2%29%7D%5B%5Cfrac%7B4.80%2A10%5E%7B-6%7DC%7D%7B0.200m%7D-%5Cfrac%7B2.26%2A10%5E%7B-6%7DC%7D%7B0.200m%7D%7D%5D%3D228391.8N%2FC)
b) for y=0.600m, r1=0.600m, r2=0.200m:
![E=\frac{1}{2\pi(8.85*10^{-12}C^2/Nm^2)}[\frac{4.80*10^{-6}C}{0.600m}-\frac{2.26*10^{-6}C}{0.200m}}]=-59345.91N/C](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B1%7D%7B2%5Cpi%288.85%2A10%5E%7B-12%7DC%5E2%2FNm%5E2%29%7D%5B%5Cfrac%7B4.80%2A10%5E%7B-6%7DC%7D%7B0.600m%7D-%5Cfrac%7B2.26%2A10%5E%7B-6%7DC%7D%7B0.200m%7D%7D%5D%3D-59345.91N%2FC)
<span>Nothing, in terms of the chemistry.
The distance between the electrodes affects the electrical resistance very slightly. Increasing the distance increases the resistance and reduces the current slightly, which reduces slightly the amount of product.
For most practical applications, for electrolysis done in a beaker, varying the distance between the electrodes will make little difference.
Increasing the concentration of the electrolyte will increase the current flow because there are more charged particles to carry charge, and increase the product yield.</span>
Answer:
V1 =8.1 m/s
Explanation:
height at highest point (h2) = 4.1 m
height at lowest point (h1) = 0.8 m
acceleration due to gravity (g) = 9.8 m/s^{2}
from conservation of energy, the total energy at the lowest point will be the same as the total energy at the highest point. therefore
mgh1 +
= mgh2 + 
where
- speed at highest point = V2
- speed at lowest point = V1
- mass of the girl and swing = m
- at the highest point, the speed is minimum (V1 = 0)
- at the lowest point the speed is maximum (V2 is the maximum speed)
- therefore the equation becomes mgh1 +
= mgh2
m(gh1 +
) = m(gh2)
gh1 +
= gh2
V1 = 
now we can substitute all required values into the equation above.
V1 = 
V1 = 
V1 =8.1 m/s
Answer:
The answer is B) 3 seconds
Explanation:
I just took the test on 2020 edge and got it right
3 kilometers, it is just 5/60 or 1/12 multiplied by 36.