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horrorfan [7]
2 years 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]2 years 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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If P = 4.00î + 3.00k, P · Q = 17.0, and P ✕ Q = −6.00ĵ, determine the vector Q.
Butoxors [25]

let the vector Q is given as

\vec Q = a\hat i + b\hat j + c\hat k

given that

P X Q = -6\hat j

here we know that

P = 4\hat i + 3 \hatk

now by above equation

(4\hat i + 3\hat k) X (a\hat i + b\hat j + c\hat k) = - 6\hat j

4b\hat k - 4c\hat j + 3a\hat j - 3b\hat i = - 6\hat j

so by comparing both sides

b = 0

4c - 3a = 6

also we know that

a^2 + b^2 + c^2 = 17^2

a^2 + 0 + (1.5 + 0.75a)^2 = 289

by solving above equation

a = 12.85 and c = 11.14

so the vector Q is given as

Q = 12.85\hat i + 11.14\hat k

8 0
4 years ago
Bohr's model of the atom included a positively charged _____ orbited by negatively charged _____. nucleus, protons nucleus, elec
alexgriva [62]
Electrons and neutrons


altho if I were you id check other sites. Hopefully I was able to help.. Have a great day :)
7 0
4 years ago
Read 2 more answers
A seesaw pivots as shown in below. What is the net torque about the pivot point?
rodikova [14]

Answer:

2.3 Nm clockwise

Explanation:

Take counterclockwise to be positive and clockwise to be negative.

∑τ = (3 N) (2.5 m) − (7 N) (1.4 m)

∑τ = 7.5 Nm − 9.8 Nm

∑τ = -2.3 Nm

The net torque is 2.3 Nm clockwise.

7 0
4 years ago
A record is spinning at the rate of 25rpm. If a ladybug is sitting 10cm from the center of the record.
marin [14]

A) Angular speed: 0.42 rev/s

B) Frequency: 0.42 Hz

C) Tangential speed: 26.4 cm/s

D) Distance travelled: 528 cm

Explanation:

A)

In this problem, the ladybug is rotating together with the record.

The angular velocity of the ladybug, which is defined as the rate of change of the angular position of the ladybug, in this problem is

\omega = 25 rpm

where here it is measured in revolutions per minute.

Keeping in mind that

1 minute = 60 seconds

We can rewrite the angular speed in revolutions per second:

\omega = 25 \frac{rev}{min} \cdot \frac{1}{60 s/min}=0.42 rev/s

B)

The relationship between angular speed and frequency of revolution for a rotational motion is given by the equation

\omega = 2 \pi f (1)

where

\omega is the angular speed

f is the frequency of revolution

For the ladybug in this problem,

\omega=0.42 rev/s

Keeping in mind that 1 rev = 2\pi rad, the angular speed can be rewritten as

\omega = 0.42 \frac{rev}{s} \cdot 2\pi = 2\pi \cdot 0.42

And re-arranginf eq.(1), we can find the frequency:

f=\frac{\omega}{2\pi}=\frac{(2\pi)0.42}{2\pi}=0.42 Hz

And the frequency is the number of complete revolutions made per second.

C)

For an object in circular motion, the tangential speed is related to the angular speed by the equation

v=\omega r

where

\omega is the angular speed

v is the tangential speed

r is the distance of the object from the axis of rotation

For the ladybug here,

\omega = 2\pi \cdot 0.42 rad/s is the angular speed

r = 10 cm = 0.10 m is the distance from the center of the record

So, its tangential speed is

v=(2\pi \cdot 0.42)(0.10)=0.264 m/s = 26.4 cm/s

D)

The tangential speed of the ladybug in this motion is constant (because the angular speed is also constant), so we can find the distance travelled using the equation for uniform motion:

d=vt

where

v is the tangential speed

t is the time elapsed

Here we have:

v = 26.4 cm/s (tangential speed)

t = 20 s

Therefoe, the distance covered by the ladybug is

d=(26.4)(20)=528 cm

Learn more about circular motion:

brainly.com/question/9575487

brainly.com/question/9329700

brainly.com/question/2506028

#LearnwithBrainly

7 0
4 years ago
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Freight trains can produce only relatively small acceleration and decelerations. (a) what is the final velocity (in m/s) of a fr
m_a_m_a [10]
(a) We will use the equation v = u + at
Initial velocity u = 5.00 m/s 
Acceleration a = 0.0600 m/s² 
time = 8 min = 8 x 60 = 480 s 
Final velocity 
= u + at 
= 5.00 + 0.0600(480) 
= 33.8 m/s

The final velocity is 33.8 m/s
5 0
3 years ago
Read 2 more answers
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