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Ket [755]
3 years ago
8

Each motor used in a continuous-duty application and rated more than one horsepower is required to be protected against overload

. A separate overload device that is responsive to motor current is one of four options permitted to protect against overload.__________
Physics
1 answer:
rosijanka [135]3 years ago
5 0

Answer:

True

Explanation:

Every electrical installation must be provided with a series of protections that make it safe, both from the point of view of the conductors and the devices connected to them, as well as the people who have to work with it.

There are many types of protections, which can make a completely safe electrical installation against any contingency, but there are three that must be used in all types of installation: lighting, domestic, power, distribution networks, auxiliary circuits, etc., already Be it low or high voltage. These three electrical protections, which we will describe in detail below are:

a) Short circuit protection.

b) Overload protection.

c) Protection against electrocution.

We understand by overloading to the excess of intensity in a circuit, due to a fault of isolation or, to a fault or excessive demand of load of the machine connected to an electric motor.

Overloads must be protected, as they can lead to the total destruction of the insulation, a network or a motor connected to it. An unprotected overload always degenerates into a short circuit.

According to the electrotechnical regulations "If the neutral conductor has the same section as the phases, the overload protection will be done with a device that only protects the phases, on the contrary if the section of the neutral conductor is inferior to that of the phases, the protection device must also control the neutral current ". In addition, a protection must be placed for each circuit derived from another main.

The most used devices for overload protection are:

- Calibrated fuses, type gT or gF (never aM)

- Circuit breakers (PIA)

- Thermal relays

For domestic circuits, lighting and small motors, the first two are usually used, as well as for short circuits, provided that the type and calibration appropriate to the circuit to be protected is used. On the contrary for three-phase motors, the so-called thermal relays are usually used.

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Which term describes the number of crests that pass a point in a given amount of time
joja [24]
The term is frequency.

The frequency is the number of vibrations per unit of time or the number of waves that passes a point per unit of time.

Every crest (and every trough) represents a pass of the wave so you can count the number of crests in an intervavl of time to find the frequency as the number of crests divided by the time elapsed. 
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3 years ago
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A man pulled a 13.0 kg object 11.8 cm vertically with his teeth. (a) How much work (in J) was done on the object by the man in t
jonny [76]

Answer:

(a)The work done by the man is -15.03J.

(b)The force exerted on the object is 127.4N.

Explanation:

Mass of the object pulled by the man is -13kg

Object is lifted 11.8 cm vertical with his teeth it means (displacement = +11.8cm = +0.118m)

Acceleration due to gravity is 9.8 \mathrm{m} / \mathrm{s}^{2}

(a) <u>Calculating the work done</u>:

Work done = mgh

Where "m" is mass of an object, "g" is acceleration due to gravity and "h" is the displacement.

\text { Work }=-13 \times 9.8 \times(+0.118 \mathrm{m})

\text { Work }=-15.03 \mathrm{J}

The work done by the man is -15.03J.

(b) <u>Calculating the force</u>:

Probably the man and the object are close to the exterior of the earth. If the rigidity required to maintained the object of consistent velocity interior the gravitational field of the earth is \mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2}

Thus the weight of the object is balanced by the force of the man's teeth on the object. That is

F = mg

\mathrm{F}=13 \times 9.8

F = 127.4N

The force exerted on the object is 127.4N.

4 0
2 years ago
3. If the distance of the screen is moved from 100. cm to 200. cm the area of light would in-crease from 150. cm^2 to
german

Answer:

3) C

4 D

5) C

Explanation:

3) given that

Initial distance of the screen = 100cm

Initial area = 150 cm^2

Final distance = 200 cm

The intensity of light is inversely proportional to the square of the distance. That is

Intensity of light I = 1/d2

And also I = P/A

1/d^2 = P/A

P = A/d^2

P1 = P2

150/100 = A/200

1.5 = A/200

A = 1.5 × 200

A = 300 cm^2

4.) Light is projected onto a screen 75.0 cm from a light source. The light intensity = 4436 lux

If the screen is moved from 75.0 cm to 150. cm, the light sensor reading will be

Using inverse square law

I = 1/d^2

I×d^2 = constant. Therefore,

4436 × 75^2 = I × 150^2

I = 24952500/22500

I = 1109 lux

5.) We can express the relationship between luminosity, brightness, and distance with a simple formula.

As we tilt the serene the area of light decreases and makes the light more concentrated.

5 0
2 years ago
When a 0.292 kg mass on a string is swung at 8.45 m/s, it feels a centripetal acceleration of 78.9 m/s^2. What is the radius of
miss Akunina [59]

Answer:

r=0.26m

Explanation:

F=mv^2/r

78.9=0.292*(8.45)^2/r

r=0.292*(8.45)^2/78.9

r=0.26m

3 0
3 years ago
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