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Evgen [1.6K]
3 years ago
5

A 06-C charge and a 07-C charge are apart at 3 m apart. What force attracts them?​

Physics
1 answer:
Vlad [161]3 years ago
5 0

Answer:

Force of 37.8 × 10^(6) N attracts the two charges

Explanation:

The force between two charges is given by

F = k*q1*q2/r²

Where q1 and q2 are 0.06 C and 0.07 C.

r is the distance between q1 and q2 which is equal to 3 m

k is a constant = 9 × 10^(9) N.m²/C²

F = (9 × 10^(9) × 0.06 × 0.07)/3²

F = 37.8 × 10^(6) N

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A 44-turn rectangular coil with length ℓ = 17.0 cm and width w = 8.10 cm is in a region with its axis initially aligned to a hor
Mumz [18]

Answer:

The maximum induced emf in the rotating coil  = 29.66V

The induced emf in the rotating coil when (t = 1.00 s) = 26.66V

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Explanation:

Lets state the parameters we are being given right from the question:

Number of rectangular coil, (N) = 44

Length of Coil, l =17cm in meters we have; (l) = 17 × 10⁻² m

Width of Coil, w =8.10cm in meters we have; (w) = 8.10 × 10⁻² m

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(a)

To calculate the induced emf in the rotating cell,we can use the formula:

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For maximum induced emf, the value of sin(ωt) will be 1

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emf_max  = NB(l × w)ω

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emf_max  = 44 × 765 × 10⁻³ ( 17 × 10⁻² × 8.10 × 10⁻² ) × 64

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(b)

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emf = NBAωsin(ωt)

substituting the parameters into the equation, we have:

emf =   44 × 765 × 10⁻³ ( 17 × 10⁻² × 8.10 × 10⁻² ) × 64 × sin (64 × 1)

=26.66V

(c)

To calculate the maximum rate of change of the magnetic flux through the rotating coil; we need to reflect on the equation for the maximum induced emf in terms of magnetic flux.

i.e emf_max = N\frac{d∅}{dt}

since emf_max = 29.66 and N = 44; we have:

29.66 =  44\frac{d∅}{dt}

\frac{d∅}{dt} = \frac{29.66}{44}

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