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balandron [24]
3 years ago
9

Your friend says that the second law of thermodynamics can't be true because life itself is a highly ordered system that wouldn'

t exist if everything tended to maximum entropy. What is your friend forgetting about life on the earth? A. Living things start out disordered and then increase in entropy. B. Living things start out ordered and then increase in entropy. C. The earth is not a closed system, and energy enters it as sunlight. D. The earth is a closed system, and thus the environment is at maximum entropy. **Not sure at all! :( Thank you!
Physics
1 answer:
spayn [35]3 years ago
8 0
This topic is actually quite controversial, but the answer in this case would be C. 

Just some food for thought, the 2nd law of thermodynamics entropy of the universe is always increasing, but that doesn't necessarily mean that earth's entropy has to. As long as the net change in entropy of the universe is increasing it doesn't matter if one planet is decreasing a nominal amount. Next, Earth as said is not a closed system and you could argue that the sunlight and energy from the sun is increasing the total energy within the system that is earth meaning that it is increasing in entropy. Next, if you consider increasing entropy as an increase in the number of possible permutations that the universe or parts of the universe can take, then it is completely possible that an ordered planet and life is possible, although rare. This theory explains why there are so many life forms and why entropy is actually increasing when divergent evolution occurs. 
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12.51 A parallel RLC circuit, which is driven by a variable frequency 2-A current source, has the following values: R = 1 kΩ, L
Anastaziya [24]

Answer:

BW = 100 rad/s

wlow = 452.49 rad/s

whigh = 552.49 rad/s

V(jwlow) =1414.21 < 45°V

V(jwhigh) =1414.21 <-45°V

Explanation:

To calculate bandwidth we have formula

BW = 1/RC

BW = 1/ 1000x10x10^¯6

BW = 100 rad/s

We will first calculate resonant frequency and quality factor for half power frequencies.

For resonant frequency

wo = 1/(SQRT LC)

wo = 1/SQRT 400×10¯³ × 10×10^¯6

wo = 500 rad/s

For Quality

Q = wo / BW

Q = 500/100

Q = 5

wlow = wo [-1/2Q+ SQRT (1/2Q)² + 1]

wlow = 500 [-1/2×5 + SQRT (1/2×5)² + 1]

wlow = 452.49 rad/s

whigh = wo [1/2Q+ SQRT (1/2Q)² + 1]

whigh = 500 [1/2×5 + SQRT (1/2×5)² + 1]

whigh = 552.49 rad/s

We will start with admittance at lower half power frequency

Y(jwlow) = (1/R) + (1/jwlow L) + (jwlow C)

Y(jwlow) = (1/1000) + (1/j×452.49×400×10¯³) + (j×452.49×10×10^¯6)

Y(jwlow) = 0.001 - j5.525×10¯³ + j4.525×10¯³

Y(jwlow) = (1-j).10¯³ S

Voltage across the network is calculated by ohm's law

V(jwlow) = I/Y(jwlow)

V(jwlow) = 2/(1-j).10¯³

V(jwlow) = 1414.2 < 45°V

Now we will calculate the admittance at higher half power frequency

Y(jwhigh) = (1/R) + (1/jwhigh L) + (jwhigh C)

Y(jwhigh) = (1/1000) + (1/j×552.49×400×10¯³) + (j×552.49×10×10^¯6)

Y(jwhigh) = 0.001 - j4.525×10¯³ + j5.525×10¯³

Y(jwhigh) = (1+j).10¯³ S

Voltage across network will be calculated by ohm's law

V(jwhigh) = I/Y(jwhigh)

V(jwhigh) = 2/(1+j).10¯³

V(jwhigh) = 1414.2 < - 45°V

6 0
4 years ago
A 41.0 g marble moving at 2.30 m/s strikes a 25.0 g marble at rest. What is the speed of each marble immediately after the colli
Gre4nikov [31]

Answer:

speed of each marble after collision will be 1.728 m/sec

Explanation:

We have given mass of the marble m_1=41gram=0.041kg

Velocity of marble v_1=2.30m/sec

Its collides with other marble of mass 25 gram

So mass of other marble m_2=25gram=0.025kg

Second marble is at so v_2=0m/sec

We have to find the velocity of second marble

From momentum conservation we know that

m_1v_1+m_2v_2=(m_!+m_2)v, here v is common velocity of both marble after collision

So 0.041\times 2.30+0.025\times 0=(0.041+0.025)v

v = 1.428 m /sec

So speed of each marble after collision will be 1.728 m/sec

6 0
3 years ago
A uniform solid disk of mass 5.00 kg and diameter 47.0 cm starts from rest and rolls without slipping down a 40.0 ∘ incline that
saveliy_v [14]

Answer:

The speed it reaches the bottom is

v=6.51m/s

Explanation:

Given: m=5.0kg, r=47cm\frac{1m}{100cm}=0.47m

Using the conservation of energy theorem

U_i=K_E+K_{ER}

m*g*h=\frac{1}{2}*m*v^2+\frac{1}{2}*I*w^2

v=r*w, I=\frac{1}{2}*m*r^2

m*g*h=\frac{1}{2}*m*(r*w)^2 +\frac{1}{2}*[\frac{1}{2} *m*r^2]*w^2

m*g*h=\frac{3}{4}*m*r^2*w^2

g*h=\frac{3}{4}*r^2*w^2

Solve to w'

w^2=\frac{4*g*h}{3*r^2}

h=x*sin(30)=6.5m*sin(30)=3.25m

w=\sqrt{\frac{4*9.8m/s^2*3.25m}{3*(0.235m)^2}}

w=27.74rad/s

v=27.74rad/s*0.235m=6.51m/s

7 0
4 years ago
A car accelerates from 300 km/h to 140 km/h in 2.53 seconds. what is the distance covered?
Snowcat [4.5K]

Answer:

Acceleration = Change in Velocity/Time

Change in Velocity = 36-18 = 18 km/h=5 m/s

Time= 5 Seconds

Acceleration = 5/5= 1 m/s2

Equation of motion,s=ut+(1/2)at2

u=18 km/h=5 m/s

t=5 s

a=1 m/s2

s= (5*5)+(1/2*1*5*5)

s=25+12.5 i.e., s=37.5 m

Hope you are clear with my explanations

7 0
3 years ago
Finish the sentence below with the word that makes the sentence true.
Keith_Richards [23]

Answer:

Animals must eat other plants or animals to obtain the <u>energy</u> stored in the food

Explanation:

One classification of living organisms, according to the way they obtain energy, is that of autotrophs and heterotrophs. The first group is represented by plants, which process their own nutrients from inorganic matter.

<u>Animals -heterotrophes- are unable to process their own nutrients</u>, so they must obtain them from other organisms, either plants or animals. External food sources provide them with nutrients, which contain the energy substrate needed to perform their vital functions.

Learn more:

Autotrophs and heterotrophs brainly.com/question/7695115

7 0
3 years ago
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