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valina [46]
3 years ago
10

A flat coil is in a uniform magnetic field. What angle between the magnetic field and the plane of the coil produces the maximum

flux, and what angle produces 90% of the maximum flux
Physics
1 answer:
Alecsey [184]3 years ago
7 0

This question is incomplete, the complete question is;

A flat coil is in a uniform magnetic field. What angle between the magnetic field and the plane of the coil produces the maximum flux, and what angle produces 90% of the maximum flux

A) max: 0°

90% of max: 90°

B) max: 90°

90% of max: 45°

C) max: 90°

90% of max: 64°

D) max: 0°

90% of max: 26°

E) max: 90°

90% of max: 80°

Answer:

Option C) max: 90°

90% of max: 64°  is the correct Answer

Explanation:

from the Suppose Area is A.

then flux at angular position O is

Ф = BAsin∅

⇒ Ф = βmaxSin∅

flux will be max when sin∅ = 1

therefore sin∅ = 1

∅ = sin⁻¹ 1

∅ = 90°

Now at 90% of max flux

Ф ⇒ 0.9βmax = βmax sin∅

0.9 = sin∅

∅ = sin⁻¹ (0.9)

∅ = 64.15° ≈ 64°

Therefore Max ∅ = 90°

90% flux = 64°

Option C) max: 90°

90% of max: 64°  is the correct Answer

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A. It does not exhibit projectile motion and follows a straight path down the ramp.

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The aqueduct passes under Johnson Road in Lancaster through a siphon. The maximum capacity of the aqueduct is 350 m3/s. The heig
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Answer:

D ≈ 8.45 m

L ≈ 100.02 m

Explanation:

Given

Q = 350 m³/s (volumetric water flow rate passing through the stretch of channel, maximum capacity of the aqueduct)

y₁ - y₂ = h = 2.00 m (the height difference from the upper to the lower channels)

x = 100.00 m (distance between the upper and the lower channels)

We assume that:

  • the upper and the lower channels are at the same pressure (the atmospheric pressure).
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  • y₁ = 2.00 m  (height of the upper channel)
  • y₂ = 0.00 m  (height of the lower channel)
  • g = 9.81 m/s²
  • ρ = 1000 Kg/m³ (density of water)

We apply Bernoulli's equation as follows between the point 1 (the upper channel) and the point 2 (the lower channel):

P₁ + (ρ*v₁²/2) + ρ*g*y₁ = P₂ + (ρ*v₂²/2) + ρ*g*y₂

Plugging the known values into the equation and simplifying we get

Patm + (1000 Kg/m³*(0 m/s)²/2) + (1000 Kg/m³)*(9.81 m/s²)*(2 m) = Patm + (1000 Kg/m³*v₂²/2) + (1000 Kg/m³)*(9.81 m/s²)*(0 m)

⇒ v₂ = 6.264 m/s

then we apply the formula

Q = v*A  ⇒   A = Q/v ⇒   A = Q/v₂

⇒   A = (350 m³/s)/(6.264 m/s)

⇒   A = 55.873 m²

then, we get the diameter of the pipe as follows

A = π*D²/4   ⇒   D = 2*√(A/π)

⇒   D = 2*√(55.873 m²/π)

⇒   D = 8.434 m ≈ 8.45 m

Now, the length of the pipe can be obtained as follows

L² = x² + h²

⇒ L² = (100.00 m)² + (2.00 m)²

⇒ L ≈ 100.02 m

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Answer:

400 N

Explanation:

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While not moving, the fricition on the refrigerator is static friction. So, \mu=0.65

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