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2 years ago

Which of the following best describes entropy

Physics

1 answer:

UkoKoshka [18]2 years ago

3 0

Answer:

Exothermic reactions increase the entropy of the surroundings. Simply put, entropy measures the dispersal of energy. Since ΔH is negative in an exothermic reaction, this must mean that ΔS will take on a positive value, indicating an increase in entropy.

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How does an inclined plane change distance and how does it change direction

ElenaW [278]

Objects are known to go down because of a unbalanced force

7 0

2 years ago

Abdominal breathing is a condition in which only the inferior half of the lungs can be seen expanding and contracting with each

lilavasa [31]

Answer:

The correct answer is - Damage to the nerves that control the diaphragm.

Explanation:

Abdominal breathing is a condition in which inferior half of the lungs can be seen relaxing or contracting and expanding with the breath. This condition occurs due to the various conditions that lead to the respiratory.

It is cause due to the damage to nerves that control the diaphragm. The phrenic nerve is one of the nerve of diaphragm initiates in the neck and passes down.

Thus, the correct answer is - Damage to the nerves that control the diaphragm.

The phrenic nerve is a nerve that originates in the neck (C3–C5) and passes down between the lung and heart to reach the diaphragm.

7 0

3 years ago

a gas has a volume of 8 liters at a temperature of 300 K the volume is then increased to 12 Liters what is the new temperature (

natali 33 [55]

300/8 = 37.5
37.5 x 12 = 450
New temp. = 450 K
Hope this helps!

6 0

3 years ago

Two metal disks, one with radius R1 = 2.45 cm and mass M1 = 0.900 kg and the other with radius R2 = 5.00 cm and mass M2 = 1.60 k

natima [27]

Answer:

part (a) $a_1\ =\ 2.9\ kg$

Part (b) $a_2\ =\ 6.25\ kg$

Explanation:

Given,

mass of the smaller disk = $M_1\ =\ 0.900\ kg$
Radius of the smaller disk = $R_1\ =\ 2.45\ cm\ =\ 0.0245\ m$

mass of the larger disk = $M_2\ =\ 1.6\ kg$
Radius of the larger disk = $R_2\ =\ 5.0\ cm\ =\ 0.05\ m$
mass of the hanging block = m = 1.60 kg

Let I be the moment of inertia of the both disk after the welding, $\therefore I\ =\ I_1\ +\ I_2\\\Rightarrow I\ =\ \dfrac{1}{2}(M_1R_1^2\ +\ M_2R_2^2)\\\Rightarrow I\ =\ 0.5\times (0.9\times 0.0245^2\ +\ 1.6\times 0.05^2)\\\Rightarrow I\ =\ 2.27\times 10^{-3}\ kgm^2$

part (a)

A block of mass m is hanging on the smaller disk,

From the f.b.d. of the block,

Let 'a' be the acceleration of the block and 'T' be the tension in the string.

$mg\ -\ T\ =\ mg\\\Rightarrow T\ =\ mg\ -\ ma\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn (1)$

Net torque on the smaller disk,

$\therefore \tau\ =\ I\alpha\\\Rightarrow TR_1\ =\ \dfrac{Ia}{R_1}\\\Rightarrow T\ =\ \dfrac{Ia}{R_1^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,enq (2)$

From eqn (1) and (2), we get,

$mg\ -\ ma\ =\ \dfrac{Ia}{R_1^2}\\\Rightarrow a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}}{0.027^2}\ +\ 1.60}\\\Rightarrow a\ =\ 2.91\ m/s^2$

part (b)

In this case the mass is rapped on the larger disk,

From the above expression of the acceleration of the block, acceleration is only depended on the radius of the rotating disk,

Let ' $a_2$ ' be the acceleration of the block in the second case,

From the above expression,

$\therefore a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}}{0.05^2}\ +\ 1.60}\\\Rightarrow a\ =\ 6.25\ m/s^2$