Answer:
Answer:
Explanation:
Given that
K=8.98755×10^9Nm²/C²
Q=0.00011C
Radius of the sphere = 5.2m
g=9.8m/s²
1. The electric field inside a conductor is zero
εΦ=qenc
εEA=qenc
net charge qenc is the algebraic sum of all the enclosed positive and negative charges, and it can be positive, negative, or zero
This surface encloses no charge, and thus qenc=0. Gauss’ law.
Since it is inside the conductor
E=0N/C
2. Since the entire charge us inside the surface, then the electric field at a distance r (5.2m) away form the surface is given as
F=kq1/r²
F=kQ/r²
F=8.98755E9×0.00011/5.2²
F=36561.78N/C
The electric field at the surface of the conductor is 36561N/C
Since the charge is positive the it is outward field
3. Given that a test charge is at 12.6m away,
Then Electric field is given as,
E=kQ/r²
E=8.98755E9 ×0.00011/12.6²
E=6227.34N/C
Give you something to compare your results with. It's always nice to be able to see what changes have been made to the original, even if it's not technically the original (I know that was worded weird, I just don't know how else to explain it.) Hope this helped!
Mechanical waves are those waves that require a material medium for propagation.
<h3>What are mechanical waves?</h3>
Generally, we define a wave as a disturbance along a medium which transfers energy. It then follows that waves move energy from one point to another.
Waves can be classified as;
- Mechanical waves
- Electromagnetic waves
Mechanical waves are those waves that require a material medium for propagation such as sound, and waves on a strings.
Learn more about mechanical waves: brainly.com/question/9242091
Only within the same technology. / / /
If both of the bulbs you're comparing are incandescent, or both fluorescent, or both CFL, or both LED, then the one that uses more power is brighter. But a CFL with the same brightness as an incandescent bulb uses less power, and an LED bulb with the same brightness as both of those uses less power than either of them.
