Answer:
the magnitude of the total angular momentum of the blades is <em>743.71 kg·m²</em>
Explanation:
Converting the angular speed into radians per second:
ω = 334 rpm · (2π rad / 1 rev) · (1 min / 60 s)
ω = 34.98 rad/s
The rotational kinetic energy of the blades is given by:
EK = 1/2 I ω²
where
- I is the moment of inertia
- ω is the angular speed
Therefore, rearranging the above equation, we get:
1/2 I ω² = EK
I ω² = 2 EK
I = 2(EK) / ω²
I = 2(4.55 × 10⁵ J) / (34.98 rad/s)²
<em>I = 743.71 kg·m²</em>
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Therefore, the magnitude of the total angular momentum of the blades is <em>743.71 kg·m²</em>.
Answer:
ΔΦ = -3.39*10^-6
Explanation:
Given:-
- The given magnetic field strength B = 0.50 gauss
- The angle between earth magnetic field and garage floor ∅ = 20 °
- The loop is rotated by 90 degree.
- The radius of the coil r = 19 cm
Find:
calculate the change in the magnetic flux δφb, in wb, through one of the loops of the coil during the rotation.
Solution:
- The change on flux ΔΦ occurs due to change in angle θ of earth's magnetic field B and the normal to circular coil.
- The strength of magnetic field B and the are of the loop A remains constant. So we have:
Φ = B*A*cos(θ)
ΔΦ = B*A*( cos(θ_1) - cos(θ_2) )
- The initial angle θ_1 between the normal to the coil and B was:
θ_1 = 90° - ∅
θ_1 = 90° - 20° = 70°
The angle θ_2 after rotation between the normal to the coil and B was:
θ_2 = ∅
θ_2 = 20°
- Hence, the change in flux can be calculated:
ΔΦ = 0.5*10^-4*π*0.19*( cos(70) - cos(20) )
ΔΦ = -3.39*10^-6
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