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Marat540 [252]
3 years ago
15

Where conventional relays, contactors and switches, which have arcing contacts, are installed in Class I, Division 1 and 2 locat

ions, they must be enclosed in ? .
Physics
1 answer:
solmaris [256]3 years ago
7 0

Answer:

They must be enclosed in explosion proof housing or general-purpose housing (as per NEC Regulations)

Explanation:

It is important to know what are Class I, Division 1 and 2 locations.

Class I: includes those industries or places where flammable gases exists and there are chances of explosions.

Division 1: Those places where hazard can happen under normal operating conditions

Division 2: Those places where hazard can happen under abnormal conditions.

The conventional relays, contactors and switches cause arcing when they operate and this is a completely normal phenomenon but with respect to these categories, it is very dangerous to operate such electrical devices without any precaution.

Therefore industries are required by the regulatory authorities to install  explosion proof housing to enclose conventional relays, contactors and switches to avoid any explosion or hazard. General-purpose housing can also be used given that it follows the regulations provided by the NEC or any other regulatory authority.

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A car travels from point A to point B, moving in the same direction but with a non-constant speed. The first half of the distanc
Dmitrij [34]

Answer:

Explanation:

From A to B

distance traveled with velocity v_1  in time t_1

\frac{d}{2}=v_1t_1----1

from B to C

distance traveled is 0.5 d with v_2  and v_3  velocity for half-half time

\frac{d}{2}=\frac{v_2t_2}{2}+\frac{v_3t_3}{2}----2

divide 1 and 2 we get

\frac{1}{1}=\frac{2v_1t_1}{v_2t_2+v_3t_3}

\frac{t_1}{t_2}=\frac{v_2+v_3}{2v_1}

Now average velocity is given by

v_{avg}=\frac{d}{t_1+t_2}

taking t_1  common

v_{avg}=\frac{2v_1t_1}{t_1(1+\frac{t_2}{t_1})}

v_{avg}=\frac{2v_1}{1+\frac{2v_1}{v_2+v_3}}

v_{avg}=\frac{2v_1(v_2+v_3)}{2v_1+v_2+v_3}  

6 0
3 years ago
The kinetic energy of a rocket is increased by a factor of eight after its engines are fired, whereas its total mass is reduced
Rudik [331]

The momentum increases by a factor of 2

Explanation:

We can solve this problem by rewriting the momentum of the rocket in terms of the kinetic energy and the mass.

The kinetic energy of the rocket is:

K=\frac{1}{2}mv^2 (1)

where

m is the mass

v is the velocity

The momentum of the rocket is

p=mv (2)

From eq.(1) we get

v=\sqrt{\frac{2K}{m}}

and substituting into (2),

p=\sqrt{2mK}

Now in this problem we have:

- The kinetic energy of the rocket is increased by a factor 8:

K' = 8K

- The mass is reduced by half:

m'=\frac{m}{2}

Substituting, we find the new momentum:

p'=\sqrt{2(\frac{m}{2}(8K)}=\sqrt{4(2mK)}=2\sqrt{2mK}=2p

So, the momentum increases by a factor of 2.

Learn more about momentum and kinetic energy:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

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#LearnwithBrainly

4 0
3 years ago
hen a constant force acts upon an object, the acceleration of the object varies inversely with its mass. When a certain constant
Sladkaya [172]

Answer:

Incomplete question

Complete question:

a constant force acts upon an object, the acceleration of the object varies inversely with its mass. When a certain constant force acts upon an object with mass 44kg , the acceleration of the object is 4m/s². If the same force acts upon another object whose mass is 11kg what is this object's acceleration?

Answer: 8m/s²

Explanation:

From the statement we deduced that acceleration varies inversely with mass where force was kept constant.

Therefore,

F/m = a or F = ma

For the first statement, substituting the mass and acceleration gives:

F = 44 x 4 = 88N

Applying the force above to the second mass gives us:

a = 88/11 = 8m/s²

3 0
3 years ago
5. You're examining some of the tiny printing on one of the newer twenty-dollar bills. A 1.5 mm tall letter appears 3 mm tall an
kow [346]

Answer:

80mm or 8cm

Explanation:

According to the lens formula,

1/f = 1/u+1/v

If the object distance u = 4cm = 40mm

Object height = 1.5mm

Image height = 3mm

First, we need to get the image distance (v) using the magnification formula Magnification = image distance/object distance = Image height/object height

v/40=3/1.5

1.5v = 120

v = 120/1.5

v = 80mm

The image distance is 80mm

To get the focal length, we will substitute the image distance and the object distance in the mirror formula to have;

1/f = 1/40+1/-80

Note that the image formed by the lens is an upright image (virtual), therefore the image distance will be negative.

Also the focal length of the converging lens is positive. Our formula will become;

1/f = 1/40-1/80

1/f = 2-1/80

1/f = 1/80

f = 80mm

The focal length of the lens 80mm or 8cm

5 0
4 years ago
Read 2 more answers
A particle at 9 AM is moving towards the east at 4 ms At 12 noon, it changes its velocity and starts moving towards the north un
nexus9112 [7]

Answer:

Explanation:

Acceleration is the time rate of change of velocity.

Acceleration and velocity are vectors

If east and north are the positive directions, the east moving vector is reduced to zero and the north moving vector increases from zero to 4 m/s.

There are 3 hours or 10800 seconds between 10 AM and 1 PM

a1 = √((-4)² + 4²) / 10800 = (√32) / 10800 m/s² ≈ 4.2 x 10⁻⁴ m/s²

There are 14400 seconds between 10 AM and 2 PM

The velocity changes are still the same

a2 = √((-4)² + 4²) / 10800 = (√32) / 14400 m/s² ≈ 3.9 x 10⁻⁴ m/s²

7 0
3 years ago
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