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Crank
3 years ago
14

A 25.0 kg object is held 8.50 m above the ground. Calculate its PE

Physics
1 answer:
pantera1 [17]3 years ago
6 0

Answer:=mgh

Explanation:

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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
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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

The maximum rate of change of the magnetic flux through the rotating coil = 0.674Wb/s

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

Magnitude of Uniform Magnetic Field (B) = 767mT= 765 × 10⁻³ T

Angular Speed of Coil, (ω) = 64 rad/s

(a)

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

emf = NBAωsin(ωt)

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

emf_max = NBAω ; if (A = l × w) , we have:

emf_max  = NB(l × w)ω

subsitituting the parameters into the above equation; we have:

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

= 29.66V

(b)

At t = 1s, the induced emf is calculated as:

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}

= 0.674 Wb/s

5 0
3 years ago
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