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Dimas [21]
4 years ago
7

While it’s impossible to design a perpetual motion machine, that is, a machine that keeps moving forever, come up with ways to k

eep some type of periodic motion going for a very long time. Explain the limits on perpetual motion. Discuss what slows down a machine and how you might minimize those effects for at least one specific form of periodic motion, such as the motion of a spring or pendulum, or circular motion. Come up with an inventive solution!
Physics
1 answer:
galben [10]4 years ago
4 0
Nope, imposible. It violates the 2nd law of thermodynamics. Usually the main culprits of the machines to not work are friction and gravity.  the rest I'll leave to you.
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Georgia [21]
It increases by a factor of 4.
6 0
3 years ago
Read 2 more answers
A two-turn circular wire loop of radius 0.63 m lies in a plane perpendicular to a uniform magnetic field of magnitude 0.219 T. I
Basile [38]

Answer:

The magnitude of the average induced emf in the wire during this time is 9.533 V.

Explanation:

Given that,

Radius r= 0.63 m

Magnetic field B= 0.219 T

Time t= 0.0572 s

We need to calculate the average induce emf in the wire during this time

Using formula of induce emf

E=-\dfrac{d\phi}{dt}

E=-B\dfrac{dA}{dt}

E=-B\dfrac{A_{2}-A_{1}}{dt}

E=B\dfrac{A_{1}-A_{2}}{dt}.....(I)

In reshaping of wire, circumstance must remain same.

We calculate the length when wire is in two loops

l=2\times 2\pi\times r_{1}

l=2\times 2\pi\times 0.63

l=7.916\ m

The length when wire is in one loop

l=2\pi\times r_{2}

7.916=2\times \pi\times r_{2}

r_{2}=\dfrac{7.916}{2\times \pi}

r_{2}=1.259\ m

We need to calculate the initial area

A_{1}=N\times\pi\times r_{1}^2

Put the value into the formula

A_{1}=2\times3.14\times(0.63)^2

A_{1}=2.49\ m^2

The final area is

A_{2}=N\times\pi\times r_{2}^2

A_{2}=1\times\pi\times(1.259)^2

A_{2}=4.98\ m^2

Put the value of initial area and final area in the equation (I)

E=0.219\dfrac{2.49-4.98}{0.0572}

E=-9.533\ V

Negative sign shows the direction of induced emf.

Hence, The magnitude of the average induced emf in the wire during this time is 9.533 V.

6 0
3 years ago
What is the fraction of the hydrogen atom's mass (11h) that is in the nucleus? the mass of proton is 1.007276 u, and the mass of
Alex_Xolod [135]

The mass fraction of the hydrogen atom is 0.9995 u.

<h3>Calculation:</h3>

We know that mass fraction can be determines as,

Mass fraction = Mass of proton/mass of H atom

Given,

Mass of proton = 1.007276 u

Mass of H atom = 1.007825 u

Put the values in the formula,

Mass fraction = Mass of proton/mass of H atom

Mass fraction = 1.007276/1.007825

                       = 0.9995 u

Hence, the mass fraction of the hydrogen atom is 0.9995 u.

Learn more about mass fraction here:

brainly.com/question/9639161

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3 0
2 years ago
A sound wave, generated at a frequency of 440 hertz has a wavelength of 2.3 meters as it travels through a solid material. The a
Jlenok [28]

The approximate speed of the sound wave traveling through the solid material is 1012m/s.

<h3>Wavelength, Frequency and Speed</h3>

Wavelength is simply the distance over which the shapes of waves are repeated. It is the spatial period of a periodic wave.

From the wavelength, frequency and speed relation,

λ = v ÷ f

Where λ is wavelength, v is velocity/speed and f is frequency.

Given the data in the question;

  • Frequency of sound wave f = 440Hz = 440s⁻¹
  • Wavelength of the wave λ = 2.3m
  • Speed of the wave v = ?

To determine the approximate speed of the wave, we substitute our given values into the expression above.

λ = v ÷ f

2.3m = v ÷ 440s⁻¹

v = 2.3m × 440s⁻¹

v = 1012ms⁻¹

v = 1012m/s

Therefore, the approximate speed of the sound wave traveling through the solid material is 1012m/s.

Learn more about Speed, Frequency and Wavelength here: brainly.com/question/27120701

8 0
2 years ago
a block of wood has a length of 4 cm a width of 5 cm and a height of 10 cm what is the volume of the wood
geniusboy [140]

Volume= Length X width X height.

Plug in the values for each and solve for the volume.

V= (L)(W)(H)

V=(4cm)(5cm)(10cm).


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