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snow_lady [41]
3 years ago
9

The total charge a battery can supply is rated in mA⋅h , the product of the current (in mA ) and the time (in h ) that the batte

ry can provide this current. A battery rated at 1000 mA⋅hr can supply a current of 1000 mA for 1.0 h , 500 mA current for 2.0 h , and so on. A typical AA rechargeable battery has a voltage of 1.2 V and a rating of 1800 mA⋅h . Part A For how long could this battery drive current through a long, thin wire of resistance 34 Ω ? Express your answer in hours. Δt Δ t = nothing h SubmitRequest Answer Provide Feedback Next
Physics
1 answer:
natita [175]3 years ago
4 0

Answer: 0.2  hours

Explanation: In order to solve this question we have to considerer that a recargeable battery can supply 1800 mA  in one hour then we have to determine how long could this battery drive current through a long, thin wire of resistance 34 Ω .

Besides, this battery has a voltage of 12 V

so by using the Ohm law we also know that V=R*I,

Fron this we can obtain:

I= V/R= 12 V/ 34 Ω=0.35 A= 350 mA

then considering that this battery can supply 1800 mA in one hour we have this battery can supply 350 mA  in x time in the form:

1hour------- 1800 mA

x hour--------350 mA

time= 350/1800= 0.2 hour

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Total displacement will be 47 meter

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E_y=1175510.2\ N.C^{-1}

The Magnitude of electric field is in the upward direction as shown directly towards the charge q_1.

Explanation:

Given:

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<u>Distance of the center from each corners</u>=\frac{1}{2} \times diagonals

diagonal=\sqrt{52.5^2+52.5^2}

diagonal=74.25\cm=0.7425\ m

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Now, electric field due to charges is given as:

E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}

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E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}

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<u>Force by each of the charges at the remaining corners:</u>

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<u> Now, net electric field in the vertical direction:</u>

E_y=E_1-E_4

E_y=1175510.2\ N.C^{-1}

<u>Now, net electric field in the horizontal direction:</u>

E_y=E_2-E_3

E_y=0\ N.C^{-1}

So the Magnitude of electric field is in the upward direction as shown directly towards the charge q_1.

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