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Scilla [17]
3 years ago
15

Name eight safety electrical devices including their functions and effects if not present.​

Engineering
1 answer:
madam [21]3 years ago
3 0

Answer:

Electricity has two hazards. A thermal hazard occurs when there is electrical overheating. A shock hazard occurs when an electric current passes through a person. Both hazards have already been discussed. Here we will concentrate on systems and devices that prevent electrical hazards. Figure 1 shows the schematic for a simple AC circuit with no safety features. This is not how power is distributed in practice. Modern household and industrial wiring require the three-wire system, shown schematically in Figure 2, which has several safety features. First is the familiar circuit breaker (or fuse) to prevent thermal overload. Second, there is a protective case around the appliance, such as a toaster or refrigerator. The case’s safety feature is that it prevents a person from touching exposed wires and coming into electrical contact with the circuit, helping prevent shocks.

Hope this helps! : )

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In a lab, scientists grew several generations of offspring of a plant using the method shown. What conclusion can you make about
exis [7]
Answer: c) they have low genetic variability among them.


When a plant is grown for several generations of offspring of a plant, then there are some common things which are to be noted which are found similar in the offspring and in the parent of the offspring. The flowers and fruits and the time or season they come in are absolutely the same.
6 0
1 year ago
An unknown immiscible liquid seeps into the bottom of an open oil tank. Some measurements indicate that the depth of the unknown
barxatty [35]

Answer:

The specific weight of unknown liquid is found to be 15 KN/m³

Explanation:

The total pressure in tank is measured to be 65 KPa in the tank. But, the total pressure will be equal to the sum of pressures due to both oil and unknown liquid.

Total Pressure = Pressure of oil + Pressure of unknown liquid

65 KPa = (Specific Weight of oil)(depth of oil) + (Specific Weight of unknown liquid)(depth of unknown liquid)

65 KN/m² = (8.5 KN/m³)(5 m) + (Specific Weight of Unknown Liquid)(1.5 m)

(Specific Weight of Unknown Liquid)(1.5 m) = 65 KN/m² - 42.5 KN/m²

(Specific Weight of Unknown Liquid) = (22.5 KN/m²)/1.5 m

<u>Specific Weight of Unknown Liquid = 15 KN/m³</u>  

4 0
3 years ago
What are the conditions for sheet generator to build up its voltage?
MA_775_DIABLO [31]

Answer:

There are six conditions

1. Poles should contain some residual flux.

2. Field and armature winding must be correctly connected so that initial mmm adds residual flux.

3. Resistance of field winding must be less than critical resistance.

4. Speed of prime mover of generator must be above critical speed.

5. Generator must be on load.

6. Brushes must have proper contact with commutators.

Explanation:

3 0
2 years ago
Explain the working and performance of a centrifugal clutch with a sketch​
FinnZ [79.3K]

Answer:

A centrifugal clutch works, as the name suggests, through centrifugal force.

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5 0
3 years ago
Read 2 more answers
An aluminum block weighing 28 kg initially at 140°C is brought into contact with a block of iron weighing 36 kg at 60°C in an in
Anika [276]

Answer:

Equilibrium Temperature is 382.71 K

Total entropy is 0.228 kJ/K

Solution:

As per the question:

Mass of the Aluminium block, M = 28 kg

Initial temperature of aluminium, T_{a} = 140^{\circ}C = 273 + 140 = 413 K

Mass of Iron block, m = 36 kg

Temperature for iron block, T_{i} = 60^{\circ}C = 273 + 60 = 333 K

At 400 k

Specific heat of Aluminium, C_{p} = 0.949\ kJ/kgK

At room temperature

Specific heat of iron, C_{p} = 0.45\ kJ/kgK

Now,

To calculate the final equilibrium temperature:

Amount of heat loss by Aluminium = Amount of heat gain by Iron

MC_{p}\Delta T = mC_{p}\Delta T

28\times 0.949(140 - T_{e}) = 36\times 0.45(T_{e} - 60)

Thus

T_{e} = 109.71^{\circ}C = 273 + 109.71 = 382.71 K

where

T_{e} = Equilibrium temperature

Now,

To calculate the changer in entropy:

\Delta s = \Delta s_{a} + \Delta s_{i}

Now,

For Aluminium:

\Delta s_{a} = MC_{p}ln\frac{T_{e}}{T_{i}}

\Delta s_{a} = 28\times 0.949\times ln\frac{382.71}{413} = - 2.025\ kJ/K

For Iron:

\Delta s_{i} = mC_{i}ln\frac{T_{e}}{T_{i}}

\Delta s_{a} = 36\times 0.45\times ln\frac{382.71}{333} = 2.253\ kJ/K

Thus

\Delta s =-2.025 + 2.253 = 0.228\ kJ/K

6 0
3 years ago
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