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disa [49]
3 years ago
5

Have researchers have now been able to cool substances to absolute zero

Physics
1 answer:
ikadub [295]3 years ago
6 0
It's a law of nature, which I don't understand too well, that we can
cool things as close to Absolute Zero as we want to, but we can
never get all the way there. 

I think that individual atoms and molecules have been cooled in
the laboratory to within a few thousandths of a Celsius degree
of it ... actually not too shabby an accomplishment !
____________________________________

WOW !  I just went and searched online for more information
on this subject.  (You can't imagine what great stuff you can find
by doing that.  You ought to try it some time.)

The 1997 Nobel Prize in Physics was awarded to a team of three
physicists who invented a method of using lasers to slow down the
motion of atoms, and that's the same thing as cooling them.  They
were able to cool some atoms to a temperature of 240 millionths
of a degree above Absolute Zero !
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Answer:

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6 0
3 years ago
Which of the following statements is true?
xenn [34]

Correct answer choice is :


C) The freezing and melting temperatures of a substance are the same.


Explanation:


Fluids have a particular temperature at which they convert into solids, identified as their freezing point. In theory, the melting point of a solid should be the same as the freezing point of the liquid. In practice, small variations among these measures can be seen. The freezing point of a matter is the same as that substance's melting point. At this distinct temperature, the substance can exist as either a solid or a liquid. At temperatures below the freezing/ melting point, the substance is a solid.


4 0
3 years ago
Read 2 more answers
A mass of 5 kg of saturated water vapor at 150 kPa is heated at constant pressure until the temperature reaches 200°C. Calculate
yulyashka [42]

Answer:

The work done by the steam is 213 kJ.

Explanation:

Given that,

Mass = 5 kg

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Using formula of work done

W=Pm\DeltaV

W=Pm(\dfrac{RT_{1}}{P_{atm}}-\dfrac{RT_{2}}{P_{atm}}

W=\dfrac{PmR}{P_{atm}}(T_{2}-T_{1})

Where,R = gas constant

T = temperature

P = pressure

P_{atm}=Atmosphere pressure

m = mass

Put the value into the formula

W=\dfrac{150\times10^{3}\times5\times287.05}{1.01\times10^{5}}\times(473-373)

W=213\ kJ

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6 0
3 years ago
A pendulum is suspended from the cusp of a cycloid cut in rigid support. The path described by the pendulum bob is cycloidal and
oksano4ka [1.4K]

Answer:

Verified that he oscillations are exactly isosynchronous with frequency ω0 = p g/l, independent of the amplitude.

Explanation:

Starting from the first principle for the derivation and to prove that the oscillations are exactly isosynchronous with frequency ω0 = p g/l, independent of the amplitude. The mathematical manipulations was applied, trigonometric identities was also applied.The steps and explanation are shown in the attachment.

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3 years ago
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