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horrorfan [7]
1 year ago
6

Recessed incandescent luminaires not marked type ic and those marked for installing directly in ___________ must not have insula

tion over the top of the luminaire.
Physics
1 answer:
slavikrds [6]1 year ago
6 0

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

#SPJ4

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A playground merry-go-round has a mass of 115 kg and a radius of 2.50 m and it is rotating with an angular velocity of 0.520 rev
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Answer:

W_f = 2.319 rad/s

Explanation:

For answer this we will use the law of the conservation of the angular momentum.

L_i = L_f

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where I_m is the moment of inertia of the merry-go-round, W_m is the initial angular velocity of the merry-go-round, I_s is the moment of inertia of the merry-go-round and the child together and W_f is the final angular velocity.

First, we will find the moment of inertia of the merry-go-round using:

I = \frac{1}{2}M_mR^2

I = \frac{1}{2}(115 kg)(2.5m)^2

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Second, we will change the initial angular velocity to rad/s as:

W = 0.520*2\pi rad/s

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Third, we will find the moment of inertia of both after the collision:

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Solving for W_f:

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