PLEASE HELP THIS IS TIMED I WILL PICK YOU THE BRAINLIEST!!!!!!!!!!!How are secondary colors of light related to the primary colo
1 answer:
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Answer:
Capacitance is 0.572×10⁻¹⁰ Farad
Explanation:
Radius = R₁ = 6.25 cm = 6.25×10⁻² m
Radius = R₂ = 15 cm = 15×10⁻² m
Dielectric constant = k = 4.8
Electric constant = ε₀ = 8.854×10⁻¹² F/m
ε/ε₀=k
ε=kε₀
∴ Capacitance is 0.572×10⁻¹⁰ Farad
Answer:
4 Ohms
Explanation
(This is seriously not as hard as it looks :)
You only need two types of calculations:
- replace two resistances, say, R1 and R2, connected in a series by a single one R. In this case the new R is a sum of the two:
- replace two resistances that are connected in parallel. In that case:
I am attaching a drawing showing the process of stepwise replacement of two resistances at a time (am using rectangles to represent a resistance). The left-most image shows the starting point, just a little bit "warped" to see it better. The two resistances (6 Ohm next to each other) are in parallel and are replaced by a single resistance (3 Ohm, see formula above) in the top middle image. Next, the two resistances (9 and 3 Ohm) are nicely in series, so they can be replaced by their sum, which is what happened going to the top right image. Finally we have two resistances in parallel and they can be replaced by a single, final, resistance as shown in the bottom right image. That (4 Ohms) is the <em>equivalent resistance</em> of the original circuit.
Using these two transformations you will be able to solve step by step any problem like this, no matter how complex.
