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3 years ago

What is entropy and how is it related to string theory?

Physics

1 answer:

balu736 [363]3 years ago

5 0

Answer:

Explanation:

String theory proposes that the fundamental constituents of the universe are one-dimensional “strings” rather than point-like particles. String theory also requires six or seven extra dimensions of space, and it contains ways of relating large extra dimensions to small ones. In statistical mechanics, entropy is an extensive property of a thermodynamic system. It quantifies the number Ω of microscopic configurations that are consistent with the macroscopic quantities that characterize the system theyre related It later developed into superstring theory, which posits a connection called supersymmetry between bosons and the class of particles called fermions. Five consistent versions of superstring theory were developed before it was conjectured in the mid-1990s that they were all different limiting cases of a single theory in 11 dimensions known as M-theory. In late 1997, theorists discovered an important relationship called the AdS/CFT correspondence, which relates string theory to another type of physical theory called a quantum field theory.

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A transformer connected to a 120-v(rms) at line is to supply 13,000V(rms) for a neon sign.

lyudmila [28]

Answer:

(a) 108

(b) 110.500 kW

Solution:

As per the question:

Voltage at primary, $V_{p} = 120\ V$ (rms voltage)

Voltage at secondary, $V_{s} = 13000\ V$ (rms voltage)

Current in the secondary, $I_{s} = 8.50\ mA$

Now,

(a) The ratio of secondary to primary turns is given by the relation:

$\frac{N_{s}}{N_{p}} = \frac{V_{s}}{V_{s}}$

where

$N_{p}$ = No. of turns in primary

$N_{s}$ = No. of turns in secondary

$\frac{N_{s}}{N_{p}} = \frac{13000}{120}$ ≈ 108

(b) The power supplied to the line is given by:

Power, P = $V_{s}I_{s} = 13000\times 8.50 = 110.500\ kW$

$\frac{V_{s}}{V_{p}} = \frac{I_{p}}{I_{s}}$

$\frac{13000}{120} = \frac{I_{p}}{8.50}$

$I_{p} = \frac{13000}{120}\times 8.50 = 920.84\ A$

6 0

3 years ago

In 2000, NASA placed a satellite in orbit around an asteroid. Consider a spherical asteroid with a mass of 5.00�1015kg and a rad

irakobra [83]

Answer:

a). $v_{1}=13.49 \frac{m}{s}$

b). $v_{2}=17.54\frac{m}{s}$

Explanation:

Using the conservation of energy and the kinetic energy of the satellite around the asteroid can model the motion in

$\frac{1}{2}*m*v^2=\frac{G*M*m}{a_{o}}$

$v^2=\frac{2*G*M*m}{a_{o}}$

$G=6.673x10^{-11}\frac{m^3}{kg*s^2}$

$M=1.20x10^{16}kg$

a).

$a_{o}=8.8km*\frac{1000m}{1km}=8800m$

$v=\sqrt{\frac{2*6.673x10^{-11}\frac{m^3}{kg*s^2}*1.2x10^{10}kg}{8800m}}$

$v_{1}=13.49 \frac{m}{s}$

b).

$a_{f}=5.2km*\frac{1000m}{1km}=5200m$

$v=\sqrt{\frac{2*6.673x10^{-11}\frac{m^3}{kg*s^2}*1.2x10^{10}kg}{5200m}}$

$v_{2}=17.54\frac{m}{s}$

3 0

3 years ago

Read 2 more answers

When an old lp turntable was revolving at 33.3 rpm, it was shut off and uniformly slowed down and stopped in 5.5 seconds. throug

velikii [3]

First, let's convert the initial angular speed from rpm (rev/min) into rad/s, keeping in mind that $1 rev = 2 \pi rad$ and 1 min=60 s:
$\omega_i = 33 \frac{rev}{min}=33\frac{rev}{min} \cdot \frac{2 \pi rad}{60 s}=3.45 rad/s$

Now we can find the angular acceleration of the lp, keeping in mind that the final speed is zero:
$\alpha = \frac{\omega_f - \omega_i}{t}= \frac{0-3.45 rad/s}{5.5 s}=-0.63 rad/s^2$
and the acceleration is negative because the LP is decelerating.

Now we can find the angle covered by the LP from the beginning to the end of its motion:
$\theta (t)= \omega_i t + \frac{1}{2}\alpha t^2 = (3.45 rad/s)(5.5 s)+ \frac{1}{2}(-0.63 rad/s^2)(5.5 s)^2=$
$=9.45 rad$

And finally, we can convert it into number of revolutions:
$1 rev : 2 \pi rad = x: 9.55 rad$
$x= \frac{9.55}{2 \pi}= 1.52 rev$

4 0

3 years ago

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stiks02 [169]

Answer:

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Explanation:

8 0

3 years ago

I'd: 9872093250, password: qqqqq, join the meeting

crimeas [40]

Why whats going on homie

3 0

3 years ago