I WILL GIVE BRAINLIEST!!!

Series circuits are characterized by the fact that there is a single pathway by which charge can travel. True or false?

Ulleksa [173]

I Think Its True My Dude Or Dudette
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Hope this helps
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Zane

6 0

4 years ago

How can having a mentor help individuals achieve their goals? A. Mentors can increase accountability. B. Mentors can increase mo

lana [24]

Your best bet is to go with D.

6 0

3 years ago

Read 2 more answers

Falling raindrops frequently develop electric charges. Does this create noticeable forces between the droplets? Suppose two 1.8

umka2103 [35]

Answer:

a) $F=2.048\times 10^{-7}\ N$

b) $a=0.1138\ m.s^{-2}$

Explanation:

Given:

mass of raindrops, $m=1.8\times 10^{-6}\ kg$

charge on the raindrops, $q=+21\times 10^{-12}\ C$

horizontal distance between the raindrops, $r=0.0044\ m$

A)

From the Coulomb's Law the force between the charges is given as:

$F=\frac{1}{4\pi.\epsilon_0} .\frac{q_1.q_2}{r^2}$

we have:

$\epsilon_0=8.854\times 10^{-12}\ C^2.N^{-1}.m^{-2}$

Now force:

$F=\frac{1}{4\pi\times 8.854\times 10^{-12}} .\frac{21\times 10^{-12}\times 21\times 10^{-12}}{0.0044^2}$

$F=2.048\times 10^{-7}\ N$

B)

Now the acceleration on the raindrops due to the electrostatic force:

$a=\frac{F}{m}$

$a=\frac{2.048\times 10^{-7}}{1.8\times 10^{-6}}$

$a=0.1138\ m.s^{-2}$

7 0

3 years ago

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When the speed of the bottle is 2 m/s, the KE is kg m2/s2. When the speed of the bottle is 3 m/s, the KE is kg m2/s2. When the s

d1i1m1o1n [39]

mass of the bottle in each case is M = 0.250 kg

now as per given speeds we can use the formula of kinetic energy to find it

1) when speed is 2 m/s

kinetic energy is given as

$K = \frac{1}{2}mv^2$

$K = \frac{1}{2}(0.250)(2)^2 = 0.5 J$

2) when speed is 3 m/s

kinetic energy is given as

$K = \frac{1}{2}mv^2$

$K = \frac{1}{2}(0.250)(3)^2 = 1.125 J$

3) when speed is 4 m/s

kinetic energy is given as

$K = \frac{1}{2}mv^2$

$K = \frac{1}{2}(0.250)(4)^2 = 2 J$

4) when speed is 5 m/s

kinetic energy is given as

$K = \frac{1}{2}mv^2$

$K = \frac{1}{2}(0.250)(5)^2 = 3.125 J$

5) when speed is 6 m/s

kinetic energy is given as

$K = \frac{1}{2}mv^2$

$K = \frac{1}{2}(0.250)(6)^2 = 4.5 J$

3 0

4 years ago

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A parallel-plate capacitor is charged by connecting it to a battery. If the battery is disconnected and then the separation betw

TEA [102]

Answer:

The charge stored in the capacitor will stay the same. However, the electric potential across the two plates will increase. (Assuming that the permittivity of the space between the two plates stays the same.)

Explanation:

The two plates of this capacitor are no longer connected to each other. As a result, there's no way for the charge on one plate to move to the other. $Q$ , the amount of charge stored in this capacitor, will stay the same.

The formula $\displaystyle Q = C\, V$ relates the electric potential across a capacitor to:

$Q$ , the charge stored in the capacitor, and
$C$ , the capacitance of this capacitor.

While $Q$ stays the same, moving the two plates apart could affect the potential $V$ by changing the capacitance $C$ of this capacitor. The formula for the capacitance of a parallel-plate capacitor is:

$\displaystyle C = \frac{\epsilon\, A}{d}$ ,

where

$\epsilon$ is the permittivity of the material between the two plates.
$A$ is the area of each of the two plates.
$d$ is the distance between the two plates.

Assume that the two plates are separated with vacuum. Moving the two plates apart will not affect the value of $\epsilon$ . Neither will that change the area of the two plates.

However, as $d$ (the distance between the two plates) increases, the value of $\displaystyle C = \frac{\epsilon\, A}{d}$ will become smaller. In other words, moving the two plates of a parallel-plate capacitor apart would reduce its capacitance.

On the other hand, the formula $\displaystyle Q = C\, V$ can be rewritten as:

$V = \displaystyle \frac{Q}{C}$ .

The value of $Q$ (charge stored in this capacitor) stays the same. As the value of $C$ becomes smaller, the value of the fraction will become larger. Hence, the electric potential across this capacitor will become larger as the two plates are moved away from one another.

3 0

3 years ago