Answer:
What would most likely happen as a result of the generator in a wind turbine breaking?
The What would most likely happen as a result of the generator in a wind turbine breaking?
The blades would not be turned.
Less steam would be produced.
Electricity would not be generated.
Solar energy would not be absorbed.
The blades would not be turned.
Less steam would be produced.
Electricity would not be generated.
Solar energy would not be absorbed.
Explanation:
F
(a) If the cornea were simply thin lens then power will be 43 diopters.
(b) This is a concave lens
The cornea is the transparent front part of the eye that covers the iris, pupil, and anterior chamber. Despite injury or disease, the cornea can still repair itself quickly. However, there are situations where damage is too severe for the cornea to heal on its own – such as with a deep injury to the cornea. The following symptoms may indicate that the cornea has sustained a substantial infection, injury or disease: Blurred vision Pain Redness.
Along with the anterior chamber and lens, the cornea refracts light, accounting for approximately two-thirds of the eye's total optical power. In humans, the refractive power of the cornea is approximately 43 diopters.
There are two types of lenses: converging and diverging and here if the cornea was simply thin then the diverging or concave lens is used in the eyes which is thin in the center than their edges.
To know more about cornea, refer: brainly.com/question/13866057
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Answer:
1.24 C
Explanation:
We know that the magnitude of the induced emf, ε = -ΔΦ/Δt where Φ = magnetic flux and t = time. Now ΔΦ = Δ(AB) = AΔB where A = area of coil and change in magnetic flux = Now ΔB = 0 - 0.750 T = -0.750 T, since the magnetic field changes from 0.750 T to 0 T.
The are , A of the circular loop is πD²/4 where D = diameter of circular loop = 16.7 cm = 16.7 × 10⁻²m
So, ε = -ΔΦ/Δt = -AΔB/Δt= -πD²/4 × -0.750 T/Δt = 0.750πD²/4Δt.
Also, the induced emf ε = iR where i = current in the coil and R = resistance of wire = ρl/A where ρ = resistivity of copper wire =1.68 × 10⁻⁸ Ωm, l = length of wire = πD and A = cross-sectional area of wire = πd²/4 where d = diameter of wire = 2.25 mm = 2.25 × 10⁻³ m.
So, ε = iR = iρl/A = iρπD/πd²/4 = 4iρD/d²
So, 4iρD/d² = 0.750πD²/4Δt.
iΔt = 0.750πD²/4 ÷ 4iρD/d²
iΔt = 0.750πD²d²/16ρ.
So the charge Q = iΔt
= 0.750π(Dd)²/16ρ
= 0.750π(16.7 × 10⁻²m 2.25 × 10⁻³ m)²/16(1.68 × 10⁻⁸ Ωm)
= 123.76 × 10⁻² C
= 1.2376 C
≅ 1.24 C
Answer:
<h3>C no.</h3>
Explanation:
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<span>The
best and most correct answer among the choices provided by the question is<span> A.
If the uncharged object is a conductor, the charged object can attract opposite
charges. </span></span>
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</span><span>Hope my answer would be a great help for you. </span> </span>
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