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docker41 [41]
2 years ago
5

"Transformer is used to change the voltage".

Engineering
1 answer:
LUCKY_DIMON [66]2 years ago
6 0

Hi! transformer is used to change the voltage through electromagnetic induction.

You might be interested in
A(n) _____________ is used commonly in open split-phase motors to disconnect the start winding from the electrical circuit when
Alenkasestr [34]

A <u>centrifugal switch</u> is used commonly in open split-phase motors to disconnect the start winding from the electrical circuit when the motor reaches approximately 75% of its rated speed.

Hope that helps!

7 0
2 years ago
For a steel alloy it has been determined that a carburizing heat treatment of 15 h duration will raise the carbon concentration
Amiraneli [1.4K]

Answer:

135 hour

Explanation:

It is given that a carburizing heat treatment of 15 hour will raise the carbon concentration by 0.35 wt% at a point of 2 mm from the surface.

We have to find the time necessary to achieve the same concentration at a 6 mm position.

we know that \frac{x_1^2}{Dt}=constant where x is distance and t is time .As the temperature is constant so D will be also constant

So \frac{x_1^2}{t}=constant

then \frac{x_1^2}{t_1}=\frac{x_2^2}{t_2} we have given x_1=2 mm\ ,t_1=15 hour\ ,x_2=6\ mm and we have to find t_2 putting all these value in equation

\frac{2^2}{15}=\frac{6^2}{t_2}

so t_2=135\ hour

5 0
3 years ago
Am I alive I really need to know?
Nesterboy [21]
Hell no,cause i’m not
5 0
3 years ago
Evaluate (204 mm)(0.004 57 kg) / (34.6 N) to three
vesna_86 [32]

Answer:

the evaluation in SI unit will be 2.69\times 10^{-5}sec^{2}

Explanation:

We have evaluate \frac{(204mm\times 0.00457kg)}{34.6N}

We know that 1 mm =10^{-3}m

So 240 mm =204\times 10^{-3}m

Newton can be written as kgm/sec^2

So \frac{(204\times 10^{-3}m)\times 0.00457kg}{34.6kgm/sec^2}=2.69\times 10^{-5}sec^{2}

So the evaluation in SI unit will be 2.69\times 10^{-5}sec^{2}

4 0
3 years ago
What is the thermal efficiency of a gas power cycle using thermal energy reservoirs at 627°C and 60°C? The thermal efficiency is
Vesna [10]

Answer

63 %

Explanation:

It is given that the reservoirs is at the temperature of 627°C and 27°C

So lower temperature that isT_L  = 60°C=273+60=333 K

And the higher temperature that is T_H  = 627°C =273+627=900 K

We know that the thermal efficiency of thermal reservoir = 1-\frac{T_L}{T_H}  = 1-\frac{333}{900} = 0.63 =63 %

So the efficiency of the reservoir is 63%

3 0
3 years ago
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