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4 years ago

What is the entropy of isolated system?

Physics

1 answer:

melamori03 [73]4 years ago

7 0

Answer: Entropy is the measure of the disorder of a system

Explanation:

Entropy is a thermodynamic quantity defined as a criterion to predict the evolution or transformation of thermodynamic systems. In addition, it is used to measure the degree of organization of a system.

In other words: Entropy is the measure of the disorder of a system and is a function of state. That is, it depends only on the state of the system.

However, in the case of an isolated system in an <u>irreversible process</u>, the value of entropy increases in the course of a process that occurs naturally. While in a <u>reversible process</u> the entropy of the isolated system remains constant.

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3 years ago

Part A If the velocity of a pitched ball has a magnitude of 47.5 m/s and the batted ball's velocity is 51.5 m/s in the opposite

nalin [4]

Explanation:

Let us assume that the mass of a pitched ball is 0.145 kg.

Initial velocity of the pitched ball, u = 47.5 m/s

Final speed of the ball, v = -51.5 m/s (in opposite direction)

We need to find the magnitude of the change in momentum of the ball and the impulse applied to it by the bat. The change in momentum of the ball is given by :

$\Delta p=m(v-u)\\\\\Delta p=0.145\times ((-51.5)-47.5)\\\\\Delta p=-14.355\ kg-m/s$

So, the magnitude of the change in momentum of the ball is 14.355 kg-m/s.

Let the the ball remains in contact with the bat for 2.00 ms. The impulse is given by :

$J=\dfrac{\Delta p}{t}\\\\J=\dfrac{14.355}{2\times 10^{-3}}\\\\J=7177.5\ kg-m/s$

Hence, this is the required solution.

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4 years ago

Just how small do you think an individual atom is

jok3333 [9.3K]

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4 years ago

Read 2 more answers

In the Rutherford model of hydrogen the electron (mass=9.11x10-31 kg) is in a planetary type orbit around the proton (mass=1.67x

Andrei [34K]

Answer:

(a) $F_g=1.62*10^{-48}N$

(b) $F_e=3.68*10^{-9}N$

Explanation:

(a) We use Newton's law of universal gravitation, in order to calculate the gravitational force between electron and proton:

$F_g=-G\frac{m_1m_2}{r^2}$

Where G is the Cavendish gravitational constant, $m_1$ and $m_2$ are the masses of the electron and the proton respectively and r is the distance between them:

$F_g=-6.67*10^{-11}\frac{N\cdot m^2}{kg^2}\frac{(9.11*10^{-31}kg)(1.67*10^{-27}kg)}{(2.5*10^{-10}m)^2}\\F_g=-1.62*10^{-48}N$

The minus sing indicates that the force is repulsive. Thus, its magnitude is:

$F_g=1.62*10^{-48}N$

(b) We use Coulomb's law, in order to calculate the electric force between electron and proton, here k is the Coulomb constant and e is the elementary charge:

$F_e=-k\frac{e^2}{r^2}\\F_e=-8.99*10^{9}\frac{N\cdot m^2}{C^2}\frac{(1.6*10^{-19}C)^2}{(2.5*10^{-10}m)^2}\\F_e=-3.68*10^{-9}N$

Its magnitude is:

$F_e=3.68*10^{-9}N$

6 0

3 years ago