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1 year ago

calculate the charge on the maximum capacitor if a battery of potential difference 1.5 volts is placed across the plates.

1 answer:

5 0

If a battery with a potential difference of 1.5 volts is placed across the plates, the maximum capacitor will have a charge of 36 V.

<h3>What possible variations are there in a 1.5 volt battery?</h3>

1 V is, by definition, a potential energy differential between two places equal to one joule for every coulomb of charge. Your query is resolved by that. Between the sites where that potential difference is measured, 1.5V denotes a potential energy differential of 1.5 joules per coulomb.

<h3>How do you determine the difference in potential energy?</h3>

ΔV=VB−VA=ΔPEq. By dividing the potential energy of a charge q that has been transported from point A to point B by the charge, we may define the potential difference between points A and B as VBVA. The joules per coulomb, sometimes known as volts (V) in honor of Alessandro Volta, are the units of potential difference.

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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

natita [175]

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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3 years ago

Two hikers are 15 miles apart and walking toward each other. They meet in 2 hours. Find the rate of each hiker if one hiker walk

Marat540 [252]

Answer:

1- 2.6 mph

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See attachment

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3 years ago

The magnetic field perpendicular to a single 16.7-cm-diameter circular loop of copper wire decreases uniformly from 0.750 T to z

sammy [17]

Answer:

1.24 C

Explanation:

We know that the magnitude of the induced emf, ε = -ΔΦ/Δt where Φ = magnetic flux and t = time. Now ΔΦ = Δ(AB) = AΔB where A = area of coil and change in magnetic flux = Now ΔB = 0 - 0.750 T = -0.750 T, since the magnetic field changes from 0.750 T to 0 T.

The are , A of the circular loop is πD²/4 where D = diameter of circular loop = 16.7 cm = 16.7 × 10⁻²m

So, ε = -ΔΦ/Δt = -AΔB/Δt= -πD²/4 × -0.750 T/Δt = 0.750πD²/4Δt.

Also, the induced emf ε = iR where i = current in the coil and R = resistance of wire = ρl/A where ρ = resistivity of copper wire =1.68 × 10⁻⁸ Ωm, l = length of wire = πD and A = cross-sectional area of wire = πd²/4 where d = diameter of wire = 2.25 mm = 2.25 × 10⁻³ m.

So, ε = iR = iρl/A = iρπD/πd²/4 = 4iρD/d²

So, 4iρD/d² = 0.750πD²/4Δt.

iΔt = 0.750πD²/4 ÷ 4iρD/d²

iΔt = 0.750πD²d²/16ρ.

So the charge Q = iΔt

= 0.750π(Dd)²/16ρ

= 0.750π(16.7 × 10⁻²m 2.25 × 10⁻³ m)²/16(1.68 × 10⁻⁸ Ωm)

= 123.76 × 10⁻² C

= 1.2376 C

≅ 1.24 C

4 0

3 years ago