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MakcuM [25]
3 years ago
9

If you disconnect the wires from the battery and then reconnect them at the opposite ends of the battery, how does that change t

he direction of the charge flow?
The direction of the charge flow is the same as the question
The direction of the charge flow still varies in different part
The direction of the charge flow is reversed

I've heard it was the first option, but my dad is adamant it is the third. How is it the first one (or third)?? Please help, been stuck on this for hours
Physics
1 answer:
IrinaVladis [17]3 years ago
3 0
Correct answer is first option: <span>The direction of the charge flow is the same as the question

Explanation:
When we have battery (or any other source of electricity) we have flow of a charge. Charge flows from positive part of battery over a conductor (most often it is a wire) towards negative part of battery. This is what always happens.

Now lets observe what we have done in this question. We had wires connected to a battery and we had a flow of charge from positive to negative pole. Then we disconnect wires and </span><span>reconnect them at the opposite ends of the battery. The question is does the direction of flow of charge change? Answer is NO. Charge still flows from positive towards negative pole. The direction of flow of charge does not depend on how the wires are connected.

Imagine this:
there is line of people stretching from one end of room to opposite end. You need to take something from front part to back part of room. Imagine people pass that object one to another. Does the direction in which this object moves depend on how this line of people is arranged? Same is for wire and charge.</span>
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Water exits a garden hose at a speed of 1.2 m/s. If the end of the garden hose is 1.5 cm in diameter and you want to make the wa
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Answer:

A 93%

Explanation:

P_1=P_2 = Pressure will be equal at inlet and outlet

\rho = Density of water = 1000 kg/m³

g = Acceleration due to gravity = 9.81 m/s²

v_1 = Velocity at inlet = 1.2 m/s

v_2 = Velocity at outlet

r_1 = Radius of inlet = \dfrac{1.5}{2}=0.75\ cm

r_2 = Radius of outlet

From Bernoulli's relation

P_1+\dfrac{1}{2}\rho v_1^2+\rho gh_1=P_2+\dfrac{1}{2}\rho v_2^2+\rho gh_2\\\Rightarrow \dfrac{1}{2}\rho v_1^2+\rho gh_1=\dfrac{1}{2}\rho v_2^2+\rho gh_2\\\Rightarrow v_2=\sqrt{2(\dfrac{1}{2}v_1^2+gh_1-gh_2)}\\\Rightarrow v_2=\sqrt{2(\dfrac{1}{2}1.2^2+9.81\times 15)}\\\Rightarrow v_2=17.19709\ m/s

From continuity equation

A_1v_1=A_2v_2\\\Rightarrow \pi r_1^2v_1=\pi r_2^2v_2\\\Rightarrow r_2=\sqrt{\dfrac{r_1^2v_1}{v_2}}\\\Rightarrow r_2=\sqrt{\dfrac{0.0075^2\times 1.2}{17.19709}}\\\Rightarrow r_2=0.00198\ m

The fraction would be

\dfrac{A_1-A_2}{A_1}\times 100=\dfrac{r_1^2-r_2^2}{r_1^2}\times 100\\ =\dfrac{0.0075^2-0.00198^2}{0.0075^2}\times 100\\ =93.0304\ \%

The fraction is 93.0304%

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