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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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A mass of 0.1 kg of helium fills a 0.2 m3 rigid tank at 350 kPa. The vessel is heated until the pressure is 700 kPa. Calculate t
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Explanation:

(a)   First, we will calculate the number of moles as follows.

       No. of moles = \frac{\text{mass}}{\text{molar mass}}

Molar mass of helium is 4 g/mol and mass is given as 0.1 kg or 100 g (as 1 kg = 1000 g).

Putting the given values into the above formula as follows.

       No. of moles = \frac{\text{mass}}{\text{molar mass}}

                             = \frac{\text{100 g}}{4 g/mol}  

                             = 25 mol

According to the ideal gas equation,

                           PV = nRT

or,       (P_{2} - P_{1})V = nR (T_{2} - T_{1})

          (6.90 atm - 3.45 atm) \times 200 L = 25 \times 0.0821 L atm/mol K \Delta T

          \Delta T = 336.17 K

Hence, temperature change will be 336.17 K.

(b)   The total amount of heat required for this process will be calculated as follows.

                   q = mC \Delta T

                      = 100 g \times 5.193 J/g K \times 336.17 K

                      = 174573.081 J/K

or,                  = 174.57 kJ/K        (as 1 kJ = 1000 J)

Therefore, the amount of total heat required is 174.57 kJ/K.

3 0
3 years ago
During a storm, the waves at this lighthouse were 5.0 meters tall top to bottom, and 10.0 m long. The waves impacted every 6.0 s
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Answer:

more info

Explanation:

7 0
2 years ago
What are the conditions under which the resultant of three coplanar forces is zero?<br>​
serg [7]
When three or more coplanar forces are acting at a point and the vector diagram closes, there is no resultant. The forces acting at the point are in equilibrium.
4 0
3 years ago
In a butcher shop, a horizontal steel bar of mass 4.94 kg and length 1.46 m is supported by two vertical wires attached to its e
mojhsa [17]

Answer:

Tension in right wire = 25.9N

Explanation:

I have attached a free body diagram to depict this question.

From the diagram, i have labelled the tensions in the strings T1 and T2.

While i labelled the weight of the bar as Wb and weight of sausage as Ws.

Now, when solving a problem like this we want to first remember that the beam is static; meaning it is not moving. From simple physics, this means that the sum of the forces in the y direction equals zero (i.e. the total downward forces equal the total upward forces)

Thus, from the diagram, the upward forces are T1 and T2 while the downward forces are Ws and Wb.

Thus;

T1 + T2 = Wb + Ws

We know that mass of bar = 4.94kg. Thus, Weight of bar(Wb) = mg = 4.94 x 9.81 = 48.46N

Also, weight of sausage (Ws) = mg = 2.49 x 9.81 = 24.43N

Thus,

T1 + T2 = 48.46N + 24.43

T1 + T2 = 72.89N - - - - - (eq 1)

Now, let's take moments about the left end of the bar.

The maximum weight of the bar will act at the centre, so distance from the Wb to left end = 1.46/2 = 0.73m

So, moments about left end;

T2 x 1.46 = (Wb x 0.73) + (Ws x 0.1)

1.46T2 = (48.46 x 0.73) + (24.43 x 0.1)

1.46T2 = 35.373 + 2.443

1.46T2 = 37.816

T2 = 37.816/1.46 = 25.9N

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3 years ago
A 2.4 mm -diameter copper wire carries a 37 A current (uniform across its cross section). Determine the magnetic field at the su
cluponka [151]

Answer:

Explanation:

We shall apply Ampere's circuital law to find out magnetic field . It is given as follows.

∫B.dl = μ₀ I , B is magnetic field , I is current ,  μ₀ is permeability .

Radius of the wire r = 1.2 x 10⁻³ m

magnetic field B will be circular in shape around the wire. If B is uniform

∫B.dl = B x 2πr  

B x 2πr  = μ₀ I

B = μ₀ I / 2πr

= 4π x 10⁻⁷ x 37 /2πx1.2 x 10⁻³

= 10⁻⁷ x 2x37 / 1.2 x 10⁻³

= 61.67 x 10⁻⁴ T

= 62  x 10⁻⁴ T

7 0
3 years ago
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