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nikitadnepr [17]
3 years ago
12

What is the suitable condition for the superconductivity and high resista (a) Weak phonon-phonon interactions (b) No interaction

s between the particles (c) Weak electron-phonon interactions (d) A strong electron-phonon interaction ос
Physics
1 answer:
Readme [11.4K]3 years ago
4 0

Answer:

(d) A strong electron-phonon interaction

Explanation:

Superconductivity -

The phenomenon of superconductivity is due to the attractive force between electrons from the exchange of the phonons that cause the bound pair of electrons known as cooper pairs .

A strong electron -phonon intercation is suitable condition for superconductivity and high resistance .

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The Lamborghini Huracan has an initial acceleration of 0.75g. Its mass, with a driver, is 1510 kg.
Cerrena [4.2K]

Answer:

11109.825 N

Explanation:

Given Data:

total mass =m=1510 kg

initial acceleration (a) =0.75g                  ( g=9.81 m/s² )

F=ma

  = (1510)*( 0.75*9.81)

  = 11109.825 N

4 0
3 years ago
A scientist directs monochromatic light toward a single slit in an opaque barrier. The light has a wavelength of 580 nm and the
vredina [299]

Answer:

a) 9.72 mm

b) 4.86 mm

Explanation:

wave length of light  λ is  580 nm = 580 \times 10⁻⁹ m

Width of slit d = 0.215\times 10⁻³ m

Distance of screen D  = 1.8 m.

Width of one fringe = \frac{\lambda\times D}{d}

Putting the values we get fringe width

= \frac{580\times10^{-9}\times1.8}{.000315}

=4.86 mm.

a) Width of central maxima = 2 times width of one fringe

= 2 times 4.86

=9.72 mm

b) width of each fringe except central fringe  is same , no matter what the order is.Only brightness changes .

So width of either of the two first order bright fringe will be same and it will be  

= 4.86 mm.

3 0
3 years ago
How can you demonstrate that charged objects exert forces, both attractive and repulsive
Tcecarenko [31]
If it attractive it has opposite pole and if it repulsive it has same pole
8 0
3 years ago
I have a bottle of gas, the bottle can expand and contract. Initially the gas is at 1 kpa of pressure and a volume of 1 Liter, a
drek231 [11]

Answer:

P₂ = 1.22 kPa

Explanation:

This problem can be solved using the equation of state:

\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}

where,

P₁ = initial pressure = 1 KPa

P₂ = final pressure = ?

V₁ = initial Volume = 1 liter

V₂ = final volume = 1.1 liter

T₁ = initial temperature = 290 k

T₂ = final temperature = 390 k

Therefore,

\frac{(1\ kPa)(1\ liter)}{290\ k} =\frac{(P_2)(1.1\ liter)}{390\ k}\\\\P_2= \frac{(1\ kPa)(1\ liter)(390\ k)}{(290\ k)(1.1\ liter)}

<u>P₂ = 1.22 kPa</u>

7 0
3 years ago
An 8.0 cm object is 40.0 cm from a concave mirror that has a focal length of 10.0 cm. Its image is 16.0 cm in front of the mirro
svet-max [94.6K]
 We can rearrange the mirror equation before plugging our values in. 
1/p = 1/f - 1/q. 
1/p = 1/10cm - 1/40cm
1/p = 4/40cm - 1/40cm = 3/40cm
40cm=3p  <-- cross multiplication
13.33cm = p

Now that we have the value of p, we can plug it into the magnification equation.

M=-16/13.33=1.2
1.2=h'/8cm
9.6=h'

So the height of the image produced by the mirror is 9.6cm.
6 0
3 years ago
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