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aev [14]
3 years ago
5

A 45.5-turn circular coil of radius 4.85 cm can be oriented in any direction in a uniform magnetic field having a magnitude of 0

.490 T. If the coil carries a current of 26.7 mA, find the magnitude of the maximum possible torque exerted on the coil.

Physics
2 answers:
il63 [147K]3 years ago
3 0

Answer:

4.399 Nm

Explanation:

The maximum Torque on a coil is given as,

τ = BNIA...................... Equation 1

Where τ = Maximum torque exerted on the coil, B = Magnetic Field, N = Number of turns, I = Current, A = Area.

Given: N = 45.5 Turns, B = 0.49 T, I = 26.7 mA = 0.0267 A,

A = πr², Where r = radius of the coil, r= 4.85 cm = 0.0485 m

A = 3.14(0.0485)²

A = 7.39×10⁻³ m².

Substitute into equation 1

τ  = 45.5×0.49×26.7×7.39×10⁻³

τ = 4.399 Nm

igomit [66]3 years ago
3 0

Explanation:

Below is an attachment containing the solution.

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In your physics lab you are given a 10.1-kg uniform rectangular plate with edge lengths 68.7 cm by 47.5 cm. Your lab instructor
VladimirAG [237]

Answer:

2.35 kgm^2

Explanation:

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Mrac [35]

Answer:

17.2 seconds

Explanation:

Given:

v₀ = 0 m/s

a₁ = 10.0 m/s²

t₁ = 3.0 s

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t₂ = 5.0 s

a₃ = -12 m/s²

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Finally, find t₃:

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The total time is:

t = t₁ + t₂ + t₃

t = 3.0 s + 5.0 s + 9.2 s

t = 17.2 s

Round as needed.

4 0
3 years ago
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