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4 years ago

Can i make an engine with magnets?

Engineering

1 answer:

Phantasy [73]4 years ago

5 0

Answer:

No, because Scientists and inventors alike have attempted to use permanent magnets alone to drive a motor. A permanent magnet motor would not produce energy and would not be a perpetual motion machine.

You might be interested in

Complete the following sentence. The skills and content of several subject areas were combined to form a new field known as a me

Svetlanka [38]

Answer:

The answer is "discipline" , I was doing the same thing a minute ago. I hope this anwser is helpful! ^^

Explanation:

5 0

3 years ago

Read 2 more answers

Need answers for these please

Step2247 [10]

Answer:

Following are given the answers one-by-one with explanation:

(a) 8 records

As records are the horizontal rows of data and we have 8 of them.

(b) 5 fields

As fields are vertical columns of data and we have 5 of them.

(c)E3000

As by sorting Makers_names in ascending order the last one would be Rany and by sorting Processor_type we have two option with Rany that are B1500 and E3000. We can see that E3000 will come at the end.

(d) Computer_type

As only two choices (Laptop and PC ) are being used under the column computer_type so it can be amended to contain Boolean data.

(e) format check

This will check that the field should contain one alphabet (capital letter) followed by 4 digits each.

i hope it will help you!

7 0

3 years ago

To measure the voltage drop across a resistor, a _______________ must be placed in _______________ with the resistor. Ammeter; S

gulaghasi [49]

Answer:

The given blanks can be filled as given below

Voltmeter must be connected in parallel

Explanation:

A voltmeter is connected in parallel to measure the voltage drop across a resistor this is because in parallel connection, current is divided in each parallel branch and voltage remains same in parallel connections.

Therefore, in order to measure the same voltage across the voltmeter as that of the voltage drop across resistor, voltmeter must be connected in parallel.

4 0

3 years ago

What advantage might there be to having the encoder located on the motor side of the gearhead instead of at the output shaft of

pickupchik [31]

The most accurate answer to that process is definitely precision. The Rotary encoder is an electro-mechanical device that converts the angular position or motion of a shaft or axle to analog or digital output signals. The efficiency of these devices is subject to the position and angle of the axis in front of the encoder.

Most cars use reduction systems in their gearboxes that convert a certain signal input into an output. Mechanically for example, a 20: 1 reduction box already infers that if there is a revolution in the input at the output there are 20. That same transferred to the encoder pulses would imply greater precision.

For example a decoder with 50 holes would have to read 1000 pulses (50 * 20) which is basically a degree of accuracy of 0.36 degrees. In this way it is possible to conclude that if the assembly of the encoder is carried out next to the motor and not at the output, it can be provided with greater precision at the time of reading.

7 0

4 years ago

Air enters a compressor operating at steady state at 1.05 bar, 300 K, with a volumetric flow rate of 21 m3/min and exits at 12 b

omeli [17]

Answer:

- 46.5171kW

Explanation:

FIrst, the value given:

P1 = 1.05 bar (Initial pressure)

P2 = 12 bar (final pressure)

Heat transfer, Q = - 3.5 kW (It is negative because the compressor losses heat to the surroundings)

Mgaseous nitrogen = Mair = 28.0134 Kg/mol (constant)

Universal gas constant, Ru = 8.3143 Kj/Kgmolk

Specific gas constant, R = 0.28699 Kj/KgK

Initial temperature, T1 = 300 K

Final temperature, T2 = 400 K

Finding the volume:

P1V1 = RT1

V1 = RT1 ÷ P1

= (0.28699 Kj/KgK X 300k) ÷ 105

Note convert bar to Kj/Nm by multiply it by 100

V1 = 0.81997 m3/Kg

To get the mass flow rate:

m = volumetric flow rate / V1

= (21 m3/min x 1/60seconds) ÷ 0.81997 m3/Kg

= 0.4268Kg/s

Using tables for the enthalpy,

hT1 = 300.19 KJ/Kg

hT2 = 400.98 KJ/Kg

The enthalpy change = hT2 - hT1

= 100.79 KJ/Kg

Power, P = Q - (m X enthalpy change)

= - 3.5 - (0.4268 X 100.79)

= - 46.5171kW

3 0

3 years ago