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bogdanovich [222]
4 years ago
6

A favorable entropy change occurs when δs is positive. what can be said about the order of the system when δs is positive?

Physics
1 answer:
iren2701 [21]4 years ago
7 0
<span>A favorable entropy change occurs when </span>δ<span>s is positive or when there is an increase in entropy. The concept of entropy tells us that isolated systems tend to go from order to disorder. That is, an increase in entropy means an increase in disorderliness. This is the natural tendency of things. </span>
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Let's practice calculating the frictional force of a skier on old "woody" skis and wet snow. The skier has a mass of 58 kg. The
kicyunya [14]

Answer:

Kinetic frictional force will be equal to 56.84 N

Explanation:

We have given mass of the skier m = 58 kg

Acceleration due to gravity g=9.8m/sec^2

Coefficient of kinetic friction \mu _k=0.1

We have to find the kinetic frictional force

Kinetic frictional force is given by

F_K=\mu _Kmg=0.1\times 58\times 9.8=56.84N

So kinetic frictional force will be equal to 56.84 N

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4 years ago
Two conducting parallel plates 5.0 × 10−3 meter apart are charged with a 12-volt potential
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Answer: 2.4×10^-3 v/m

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Potential difference between plates (v) = 12v

Force on electronic charge (f) = 3.8×10^-16 N

Strength of electric field (E) =?

The formulae that relates potential difference, eoectiic field strength and distance between plates is given as

v = Ed

By substituting the parameters, we have that

12 = E × 5.0×10^-3

E = 12/ 5.0 × 10^-3

E = 2.4×10^-3 v/m

7 0
3 years ago
____, one of Saturn's icy moons, is unusual in the solar system in that it has volcanic activity that ejects plumes of icy parti
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Answer: Enceladus

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The mass of a planet is 3.7 x 1024 kg. If the planet has a radius of 9.2 x 106 m what is the acceleration of gravity for a perso
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Explanation:

It s given that,

Mass of a planet, M=3.7\times 10^{24}\ kg

Radius of a planet, R=9.2\times 10^{6}\ m

(1) We need to find the acceleration due to gravity for a person on the surface of the planet. Its formula is given by :

g=\dfrac{GM}{R^2}

g=\dfrac{6.67\times 10^{-11}\ Nm^2/kg^2\times 3.7\times 10^{24}\ kg}{(9.2\times 10^{6}\ m)^2}

g=2.91\ m/s^2

(2) The escape velocity is given by :

v=\sqrt{\dfrac{2GM}{R}}

v=\sqrt{{\dfrac{2\times 6.67\times 10^{-11}\ Nm^2/kg^2\times 3.7\times 10^{24}\ kg}{9.2\times 10^{6}\ m}}

v = 7324.61 m/s

Hence, this is the required solution.

3 0
3 years ago
How are the spiral arms of the milky way detected?
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The spiral structure emerges when galactic clusters (open), H II regions and O & B type stars (young stars) are used as tracers. We know this to be true as other pinwheel galaxies exhibit the same patterns across these tracers as in the milky way.
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4 years ago
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