It is itself. This question does not make sense.
Answer:
Recessed incandescent luminaires not marked type ic and those marked for installing directly in insulated ceilings must not have insulation over the top of the luminaire.
Explanation:
Depending on how they interact with insulation, lighting fixtures are rated at various levels. Non-IC rated lighting fixtures can accommodate higher wattage bulbs, but they also pose the greatest fire risk when used with the incorrect insulation.
In locations with insulation, light fixtures that are not IC rated may be installed. But there is a condition. The distance between the fixture and any insulation should be 3 inches. But the 3 inch gap in the insulation would negate the goal of insulation by producing a lot of uninsulated space, so this defies logic. Building a box-style cover to cover the fixture on the attic side is one option to fix this. Drywall or foil-faced foam insulation can be used to create this box. After the cover is put in place, insulation can be added for maximum effectiveness.
To learn more about recessed incandescent luminaries. Click brainly.com/question/17218799
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Answer:
(a) 42.28°
(b) 37.08°
Explanation:
From the principle of refraction of light, when light wave travels from one medium to another medium, we have:
= sinθ
/sinθ
In the given problem, we are given the refractive indices of light which are parallel and perpendicular to the axis of the optical lens as 1.4864 and 1.6584 respectively.
For critical angle θ
= θ
, θ
= 90°; 
(a) 
= sinθ
/sin90°
0.6728 = sinθ![_{c}θ[tex]_{c} = sin^(-1) 0.6728 = 42.28°(b) [tex]n_{a} = 1.6584](https://tex.z-dn.net/?f=_%7Bc%7D%3C%2Fp%3E%3Cp%3E%CE%B8%5Btex%5D_%7Bc%7D%20%3D%20sin%5E%28-1%29%200.6728%20%3D%2042.28%C2%B0%3C%2Fp%3E%3Cp%3E%3C%2Fp%3E%3Cp%3E%28b%29%20%5Btex%5Dn_%7Ba%7D%20%3D%201.6584)
= sinθ
/sin90°
0.60299 = sinθ[tex]_{c}
θ[tex]_{c} = sin^(-1) 0.60299 = 37.08°
The surface of the earth changes. Some changes are due to slow processes, such as erosion and weathering, and some changes are due to rapid processes, such as landslides, volcanic eruptions, and earthquakes.
Orient the semi-circle arc such that it is symmetric with respect to the y-axis. Now, by symmetry, the electric field in the x-direction cancels to zero. So the only thing of interest is the electric field in the y-direction.
dEy=kp/r^2*sin(a) where k is coulombs constant p is the charge density r is the radius of the arc and a is the angular position of each point on the arc (ranging from 0 to pi. Integrating this renders 2kq/(pi*r^3). Where k is 9*10^9, q is 9.8 uC r is .093 m
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