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1 year ago

What precautions should be taken to avoid the overloading of domestic electric circuits.

Engineering

1 answer:

madam [21]1 year ago

4 0

The precautions that should be taken to avoid the overloading of domestic electric circuits are:

Do not put high voltage wires in one socket.
Do not use many electric appliances of high power at the same time.

<h3>What are electric circuits?</h3>

Electric circuits are wires or devices that give electricity to devices that run on electricity. Running of electric devices should be done carefully because our body can come in contact with the current.

Thus, the precautions are to keep high voltage lines away from one socket. Use only a few high-power electric appliances at once.

To learn more about electric circuits, refer to the below link:

brainly.com/question/28221759

#SPJ4

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A 3-m wide rectangular channel has a flow velocity of 1.8 m/s when the depth of flow is 1.2 m. what will be the flow velocity wh

valina [46]

Answer:

The flow velocity reduces to 0.72 m/s

Explanation:

According to the equation of continuity discharge in the channel should remain same

Thus we have

$A_{1}V_{1}=A_{2}V_{2}$

For a rectangular channel we have $Area=Depth\times Width$

Applying values in the continuity equation and since the width of the channel remains constant 3.0 m we have

$3\times 1.8\times 1.2=3\times 3\times V_{2}\\\\\therefore V_{2}=\frac{6.48}{9}=0.72m/s$

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2 years ago

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Yes I believe so.......

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3 years ago

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Neporo4naja [7]

Answer:

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2 years ago

If the bolt head and the supporting bracket are made of the same material having a failure shear stress of 'Tra;i = 120 MPa, det

Nina [5.8K]

Answer:

P=361.91 KN

Explanation:

given data:

brackets and head of the screw are made of material with T_fail=120 Mpa

safety factor is F.S=2.5

maximum value of force P=??

solution:

to find the shear stress

T_allow=T_fail/F.S

=120 Mpa/2.5

=48 Mpa

we know that,

V=P

Area for shear head:

A(head)=π×d×t

=π×0.04×0.075

=0.003×πm^2

Area for plate:

A(plate)=π×d×t

=π×0.08×0.03

=0.0024×πm^2

now we have to find shear stress for both head and plate

For head:

T_allow=V/A(head)

48 Mpa=P/0.003×π ..(V=P)

P =48 Mpa×0.003×π

=452.16 KN

For plate:

T_allow=V/A(plate)

48 Mpa=P/0.0024×π ..(V=P)

P =48 Mpa×0.0024×π

=361.91 KN

the boundary load is obtained as the minimum value of force P for all three cases. so the solution is

P=361.91 KN

note:

find the attached pic

7 0

3 years ago

Re armature of a 4 pole DC generator is required to generate an emf of 520v on open circuit when revolving at a speed of 660rpm.

bija089 [108]

Since the armature is wave wound, the magnetic flux per pole is 0.0274 Weber.

Given the following data:

Emf = 520 Volts

Speed = 660 r.p.m

Number of armature conductors = 144 slots

Number of poles = 4 poles

Number of parallel paths = 2

To find the magnetic flux per pole:

Mathematically, the emf generated by a DC generator is given by the formula;

$E = \frac{\theta ZN}{60}$ × $\frac{P}{A}$

Where:

E is the electromotive force in the DC generator.

Z is the total number of armature conductors.

N is the speed or armature rotation in r.p.m.

P is the number of poles.

A is the number of parallel paths in armature.

Ф is the magnetic flux.

First of all, we would determine the total number of armature conductors:

$Z = 144$ × $2$ × $3$

Z = 864

Substituting the given parameters into the formula, we have;

$520 = \frac{\theta (864)(660)}{60}$ × $\frac{4}{2}$

$520 = \theta (864)(11)$ × $2$

$520 = 19008 \theta \\\\\Theta = \frac{520}{19008}$

Magnetic flux = 0.0274 Weber.

Therefore, the magnetic flux per pole is 0.0274 Weber.

5 0

2 years ago