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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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BRAINLIEST
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Answer:

Explanation:

1) acceleration is the change in velocity of a body with respect to time.

acceleration = velocity/time

Given

velocity = 139m/s

time = 20secs

acceleration = 139/20

acceleration = 6.95m/s²

Hence its average acceleration during the first 20 seconds of the launch is 6.95m/s²

2) Speed is the rate of change of distance with respect to time.

average speed = distance/time

Time = distance/speed

Time = 100/7823

Time = 0.013s

3) Using the equation of motion v² = u²+2as

v² = 0²+2(15)(60)

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7 0
3 years ago
The Ha line of the Balmer series is emitted in the transition from n-3 to n 2. Compute the wavelength of this line for H and 2H.
Citrus2011 [14]

Explanation:

According to Rydberg's formula, the wavelength of the balmer series is given by:

\frac{1}{\lambda}=R(\frac{1}{2^2}-\frac{1}{3^2})

R is Rydberg constant for an especific hydrogen-like atom, we may calculate R for hydrogen and deuterium atoms from:

R=\frac{R_{\infty}}{(1+\frac{m_e}{M})}

Here, R_{\infty} is the "general" Rydberg constant, m_e is electron's mass and M is the mass of the atom nucleus

For hydrogen, we have, M=1.67*10^{-27}kg:

R_H=\frac{1.09737*10^7m^{-1}}{(1+\frac{9.11*10^{-31}kg}{1.67*10^{-27}kg})}\\R_H=1.09677*10^7m^{-1}

Now, we calculate the wavelength for hydrogen:

\frac{1}{\lambda}=R_H(\frac{1}{2^2}-\frac{1}{3^2})\\\lambda=[R_H(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=[1.0967*10^7m^{-1}(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=6.5646*10^{-7}m=656.46nm

For deuterium, we have M=2(1.67*10^{-27}kg):

R_D=\frac{1.09737*10^7m^{-1}}{(1+\frac{9.11*10^{-31}kg}{2*1.67*10^{-27}kg})}\\R_D=1.09707*10^7m^{-1}\\\\\lambda=[R_D(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=[1.09707*10^7m^{-1}(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=6.5629*10^{-7}=656.29nm

5 0
3 years ago
3. A concrete highway is built of slabs 12 m long (20o C). How wide should the expansion cracks between the slabs be (at 20o C)
lesantik [10]

Answer:

\Delta x=2.304\times 10^{-2}\ m=2.304\ cm is the minimum gap between the slabs

Explanation:

Given:

  • length of the concrete highway, l=12\ m
  • coefficient of thermal expansion, \alpha=12\times 10^{-6}\ ^{\circ}C^{-1}
  • range of temperature variation, (-30^{\circ}C\ to\ 50^{\circ}C) \Rightarrow \Delta T=80^{\circ}C

<u>Now form the equation of thermal expansion:</u>

\Delta l=l\times \alpha\times \Delta T

\Delta l=12\times (12\times 10^{-6})\times 80

\Delta l=1.152\times 10^{-2}\ m

Since each slab of the highway expands by the above length so the minimum gap between the slabs to prevent buckling:

\Delta x=2\times \Delta l

\Delta x=2\times( 1.152\times 10^{-2})

\Delta x=2.304\times 10^{-2}\ m=2.304\ cm is the minimum gap between the slabs

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