All categories

3 years ago

He pictures show a grassy field that has been abandoned. Which best explains why this is an example of increasing entropy?

Physics

1 answer:

solmaris [256]3 years ago

6 0

<h2>Answer:</h2>

The correct answer is option D. Which is "Over time, the lawn has naturally become disorderly".

<h3>Explanation:</h3>

Entropy, the measure of a system's thermal energy per unit temperature that is unavailable for doing useful work.
It is also a measure of the molecular disorder, or randomness, of a system.
According to above definitions of entropy option D is correct.
<u>So entropy of system is randomness/disorder that is increasing with time in case of lawn.</u>

You might be interested in

You have a small piece of iron at 25 °C and place it into a large container of water at 75 °C. Which of these could be the tempe

Ivanshal [37]

50 degrees because the would most likely equal out

8 0

3 years ago

A horizontal clothesline is tied between 2 poles, 12 meters apart. When a mass of 1 kilograms is tied to the middle of the cloth

nadya68 [22]

Answer:

The tension on the clotheslines is $T = 8.83 \ N$

Explanation:

The diagram illustrating this question is shown on the first uploaded image

From the question we are told that

The distance between the two poles is $d = 12 \ m$

The mass tie to the middle of the clotheslines $m = 1 \ kg$

The length at which the clotheslines sags is $l = 4 \ m$

Generally the weight due to gravity at the middle of the clotheslines is mathematically represented as

$W = mg$

let the angle which the tension on the clotheslines makes with the horizontal be $\theta$ which mathematically evaluated using the SOHCAHTOA as follows

$Tan \theta = \frac{ 4}{6}$

=> $\theta = tan^{-1}[\frac{4}{6} ]$

=> $\theta = 33.70^o$

So the vertical component of this tension is mathematically represented a

$T_y = 2* Tsin \theta$

Now at equilibrium the net horizontal force is zero which implies that

$T_y - mg = 0$

=> $T sin \theta - mg = 0$

substituting values

$T = \frac{m*g}{sin (\theta )}$

substituting values

$T = \frac{1 *9.8}{2 * sin (33.70 )}$

$T = 8.83 \ N$

6 0

3 years ago

An ideal Diesel cycle has a compression ratio of 18 and a cutoff ratio of 1.5. Determine the (1) maximum air temperature and (2)

weqwewe [10]

Answer:

(1) The maximum air temperature is 1383.002 K

(2) The rate of heat addition is 215.5 kW

Explanation:

T₁ = 17 + 273.15 = 290.15

$\frac{T_2}{T_1} =r_v^{k - 1} =18^{0.4} =3.17767$

T₂ = 290.15 × 3.17767 = 922.00139

$\frac{T_3}{T_2} =\frac{v_3}{v_2} = r_c = 1.5$

Therefore,

T₃ = T₂×1.5 = 922.00139 × 1.5 = 1383.002 K

The maximum air temperature = T₃ = 1383.002 K

(2)

$\frac{v_4}{v_3} =\frac{v_4}{v_2} \times \frac{v_2}{v_3} = \frac{v_1}{v_2} \times \frac{v_2}{v_3} = 18 \times \frac{1}{1.5} = 12$

$\frac{T_3}{T_4} =(\frac{v_4}{v_3} )^{k-1} = 12^{0.4} = 2.702$

Therefore;

$T_4 = \frac{1383.002}{2.702} =511.859 \ k$

$Q_1 = c_p(T_3-T_2)$

Q₁ = 1.005(1383.002 - 922.00139) = 463.306 kJ/jg

Heat rejected per kilogram is given by the following relation;

$c_v(T_4-T_1)$ = 0.718×(511.859 - 290.15) = 159.187 kJ/kg

The efficiency is given by the following relation;

$\eta = 1-\frac{\beta ^{k}-1}{\left (\beta -1 \right )r_{v}^{k-1}}$

Where:

β = Cut off ratio

Plugging in the values, we get;

$\eta = 1-\frac{1.5 ^{1.4}-1}{\left (1.5 -1 \right )18^{1.4-1}}= 0.5191$

Therefore;

$\eta = \frac{\sum Q}{Q_1}$

$\therefore 0.5191 = \frac{150}{Q_1}$

Heat supplied = $\frac{150}{0.5191} = 288.978 \ hp$

Therefore, heat supplied = 215491.064 W

Heat supplied ≈ 215.5 kW

The rate of heat addition = 215.5 kW.

7 0

4 years ago

The mertric unit of electrical power is the _____. A. watt B.volt C.ampere D.ohm

sladkih [1.3K]

Answer:

A - Watt

Explanation:

Watt is the unit of electrical power in a metric system, expressed in terms of energy per second, equal to the work done at a rate of 1 joule per second is

power formula is P = V × I,

where V is the voltage in a circuit

I is the current flowing through that circuit.

In the SI (metric) system, the units of power are watts.

5 0

3 years ago

A 2kg object is moving horizontally with a speed of 4m/s 2. How much net force is required t keep the object moving at this spee

dezoksy [38]

To continue moving at constant speed in a straight line requires NO net force. Zero. Nada. If there IS any net force on the object, then its speed or direction will change.

8 0

3 years ago