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jarptica [38.1K]
3 years ago
13

I'm in  my lab, writing an essay about the properties of Bose Einstein Condensate. I forgot the formula of producing the condens

ate. Can someone help me?
Physics
1 answer:
Artyom0805 [142]3 years ago
7 0
This is a big looking question for school and for 5 points.It seems to relate to superfluidity in helium III and helium IV and something called the "lambda" point. Though I can't do it justice yet and here, it looks as though i can find tis in "Heat and Thermodynamics", Mark W Zemansky ...If rho is the density of helium II, rhon the density of the normal part, and rhoz the density of the superfluid part rho=rhon+rhoz.At the lambda point, all the atoms are normal and rhon/rho = 1, whereas at absolute zero all the atoms are superfluid and rhon/rho=0.Best I can do with the info and point available ..ps, I think that both Bose and Einstein were nobel prize winners, and the word "Boson" is in honour of the, I believe, Indian physicist Bose. It is a very interesting question, and I can in no way do it justice here.
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Determine a formula for the magnitude of the force F exerted on the large block (Mc) so that the mass Ma does not move relative
SVEN [57.7K]

Answer:

The magnitude of the force F is given by

F =  (M_{a} + M_{b} + M_{c} ) *(M_{b}*g/(\sqrt{M_{a} ^{2}-M_{b} ^{2}}))

Explanation:

Given there are three blocks of masses M_{a}, M_{b} and M_{c} (ref image in attachment)

When all three masses move together at an acceleration a, the force F is given by

F =  (M_{a} + M_{b} + M_{c} ) *a    ................(equation 1)

Also it is given that M_{a} does not move with respect to M_{c}, which gives tension T  is exerted on pulley  by M_{a} only, Hence tension T is

T = M_{a} *a    ..........(equation 2)

There is also also tension exerted by M_{b}. There are two components here: horizontal due to acceleration a and vertical component due to gravity g. Thus tension is given by

T = M_{b} \sqrt{a^{2} +g^{2} }   ................(equation 3)

From equation 2 and 3, we get

M_{a} *a  = M_{b} \sqrt{a^{2} +g^{2} }  

Squaring both sides we get

M_{a} ^{2} *a^{2} = M_{b} ^{2} * (a^{2}+g^{2})

M_{a} ^{2} *a^{2} = (M_{b} ^{2} * a^{2})+ (M_{b} ^{2} *g^{2})

(M_{a} ^{2}  -  M_{b} ^{2}) * a^{2} = M_{b} ^{2} *g^{2}

a^{2} = M_{b} ^{2} *g^{2}/(M_{a} ^{2}  -  M_{b} ^{2})

Taking square root on both sides, we get acceleration a

a = M_{b}*g/(\sqrt{M_{a} ^{2}-M_{b} ^{2}})

Hence substituting the value of a in equation 1, we get

F =  (M_{a} + M_{b} + M_{c} ) *(M_{b}*g/(\sqrt{M_{a} ^{2}-M_{b} ^{2}}))

3 0
3 years ago
Asteroids are between 1000 km and less than 10 m in diameter. What is the diameter of most asteroids?
e-lub [12.9K]

This question needs research to be answered. From the given information alone it can't be answered without making wild assumptions.

Ideally, you need to take a look at a distribution (or a histogram) of asteroid diameters, identify the "mode" of such a distribution, and find the corresponding diameter. That value will be the answer.

I am attaching one such histogram on asteroid diameters from the IRAS asteroid catalog I could find online. (In order to get a single histogram, you need to add the individual curves in the figure first). Eyeballing this sample, I'd say the mode is somewhere around 10km, so the answer would be: the diameter of most asteroid from the IRAS asteroid catalog is about 10km.

7 0
3 years ago
The pulley shown in the attached diagram has a diameter of 30 centimeters and a mass of 19 kilograms. The pulley is a solid disk
aleksandrvk [35]

Answer:

Explanation:

a) I = ½mR² = ½(19)(0.15²) = 0.21375 kg•m²

b) τ = Fnet(r) = (25 - 12)(0.15) = 1.95 N•m

c) CCW

d) a = τ/I = 1.95 / 0.21375 = 9.12280701... = 9.1 rad/s²

e) CCW

4 0
2 years ago
what are the 3 properties of components of the universe that can be determined using electromagnetic radiation?
stepan [7]
Your answer is electricity, light and magnetism.  They can be determined usinf elecromagnetic radioation. 
<span>
Even the energy  can't be detected by our eyes, there are a lot of measurement  instruments that can measure infrared (IR), gamma rays, radio or X-rays or ultraviolet (UV)</span>
4 0
3 years ago
A 30-mm-diameter copper rod is 1 m long with a yield strength of 70 MPa. Determine the axial force necessary to cause the diamet
ivolga24 [154]

Explanation:

Given data:

d = 30 mm = 0.03 m

L = 1m

S_{y} = 70 Mpa

Δd = -0.0001d

Axial force = ?

validity of elastic deformation assumption.

Solution:

O'₂ = Δd/d = (-0.0001d)/d = -0.0001

For copper,

v = 0.326      E = 119×10³ Mpa

O'₁ = O'₂/v = (-0.0001)/0.326 = 306×10⁶

∵δ = F.L/E.A    and σ = F/A so,

σ = δ.E/L = O'₁ .E = (306×10⁻⁶).(119×10³) = 36.5 MPa

F = σ . A = (36.5 × 10⁻⁶) . (π/4 × (0.03)²) = 25800 KN

S_{y} = 70 MPa > σ = 36.5 MPa

∵ elastic deformation assumption is valid.

so the answer is

F = 25800 K N            and     S_{y} > σ

3 0
3 years ago
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