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Marrrta [24]
2 years ago
6

How much charge does a 9.0 v battery transfer from the negative to the positive terminal while doing 45 j of work?

Physics
1 answer:
bija089 [108]2 years ago
6 0

Here is your answer

5 coulomb

REASON :

We know that

Potential difference, V= W/q

where, W is work done

and, q is magnitude of charge

Given,

V= 9.0 v and W= 45 J

So,

using above relation, we get

9= 45/q

q= 45/9

q= 5 coulomb

HOPE IT IS USEFUL

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Answer:

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3 years ago
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A 123-turn circular coil of radius 2.41 cm is immersed in a uniform magnetic field that is perpendicular to the plane of the coi
klasskru [66]

Answer:

67.44 V

Explanation:

Number of turns N =123

Radius = 2.41 cm =0.0241 m

The magnetic field strength increases from 50.9 T to 96.3 T so change in magnetic field dB = 96.3-50.9=45.4 T

Time interval dt = 0.151 sec

We know that the induced emf e=-NA\frac{dB}{dt}

Area A=\pi r^2=3.14\times 0.0241^2=1.8237\times 10^{-3}m^2

Putting all these values in emf equation e=-123\times 1.8237\times 10^{-3}\times \frac{45.4}{0.151}=-67.44\ V here negative sign indicates that it opposes the cause due to which it is produced

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What units for time must be used in calculating power
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Explanation:

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A ball is tied to the end of a cable of negligible mass. The ball is spun in a circle with a radius 2.00 m making 0.700 revoluti
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INT Raindrops acquire an electric charge as they fall. Suppose a 2.0-mm-diameter drop has a charge of 12 pC; these are both very
horsena [70]

Answer:

W = 2.3 10² F_{e}

Explanation:

The force of the weight is

         W = m g

         

let's use the concept of density

         ρ= m / v

the volume of a sphere is

         V = \frac{4}{3} π r³

         V = \frac{4}{3} π (1.0 10⁻³)³

         V = 4.1887 10⁻⁹ m³

the density of water ρ = 1000 kg / m³

          m = ρ V

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now let's look for the electric force

           F_e = q E

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           \frac{W}{F_e} = 2,281 10²

             

             W = 2.3 10² F_{e}

therefore the weight of the drop is much greater than the electric force

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