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Anton [14]
4 years ago
13

Follow the path of the current leaving the battery and returning back to the battery through two light bulbs if they are a) conn

ected in series and b) connected in parallel. Please explain these are all the points i have left
Physics
2 answers:
miskamm [114]4 years ago
7 0
Well there's only one path for the current to travel through in a series circuit.
there are multiple paths in a parallel circuit.    
OleMash [197]4 years ago
6 0

Answer:

If we have only one path, this means that the circuit is connected in series, one light bulb after the other, the current decreases a bit of intensity when it passes through the first light bulb and reaches the other with less intensity, so in this case, you may see that the first one brights more intensely than the second one, if there is a remanent of current, it goes again to the battery. In the parallel case, we have a bifurcation of the path, where the current divides into both paths and each part goes to a single light bulb, here we will see both light bulbs shine with the same intensity.

so the paths are:

Series:

battery----light bulb--- light bulb---battery

Parallel:

               light bulb

battery---<                 >-------battery

                light bulb

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8 0
3 years ago
Un reloj de péndulo de largo L y período T, aumenta su largo en ΔL (ΔL &lt;&lt; L). Demuestre que su período aumenta en: ΔT = π
Kruka [31]

Answer:

 ΔT = \pi \ \frac{\Delta L}{\sqrt{Lg} }

Explanation:

In a simple harmonic motion, specifically in the simple pendulum, the angular velocity

          w = \sqrt{\frac{g}{L} }

angular velocity and period are related

          w = 2π / T

we substitute

          2π / T = \sqrt{\frac{g}{L} }

          T = 2\pi  \ \sqrt{\frac{L}{g} }

In this exercise indicate that for a long Lo the period is To, then and increase the long

          L = L₀ + ΔL

we substitute

           T = 2\pi  \ \sqrt{\frac{L + \Delta L}{g} }

            T = 2\pi  \ \sqrt{\frac{L}{g} } \ \sqrt{1+ \frac{\Delta L}{L} }

in general the length increments are small ΔL/L «1, let's use a series expansion

           \sqrt{1+ \ \frac{\Delta L}{L} } = 1 + \frac{1}{2} \frac{\Delta L}{L} + ...  

we keep the linear term, let's substitute

           T = 2\pi  \ \sqrt{\frac{L}{g} } \ ( 1 + \frac{1}{2} \frac{\Delta L}{L}  )  

if we do

           T = T₀ + ΔT

           

           T₀ + ΔT = 2\pi  \sqrt{\frac{\Delta L}{g} }  + \pi  \ \sqrt{\frac{L}{g} } \ \frac{\Delta L}{L}

            T₀ + ΔT = T₀ + \pi  \sqrt{\frac{1}{Lg} } \ \Delta L

            ΔT = \pi \ \frac{\Delta L}{\sqrt{Lg} }

4 0
3 years ago
What is the acceleration of an object going from O m/s to 25 m/s in 5s?
kobusy [5.1K]

Answer:

5m/s^2 is the acceleration.

8 0
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OleMash [197]

Answer:

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algol13

Their atoms have eight electrons in their valence shells, so noble gases are very unreactive.

Explanation:

The octet rule state that atoms tend to complete their last energy levels with eight electrons, and that this configuration make them very stable and unreactive.

Noble gases are characterized as unreactive atoms, and this is associated with the fact that they have a complete valence shell, it means that they have eight electrons on it (they follow the octet rule).

Atoms with less electrons on their valence shells tend to react with another atom, forming bonds, to complete their valence shells (with eight electrons).

<h2>---</h2>

Hope this helps you! Have a great day and or night love <3

3 0
1 year ago
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