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galben [10]
3 years ago
5

Zoe went on a fishing trip. On the first day she caught 9 catfish and 6 trout. On second caught 17 catfish and 13 trought. On wh

ich day of the fishing trip did zoe catch a lower ratio of catfish to trout?
Physics
1 answer:
denis23 [38]3 years ago
4 0

Answer:

<em>Zoe caught a lower ratio of catfish to trout on the second day of her fishing trip</em>

Explanation:

<u>Proportions and Ratios</u>

To compare two quantities we can perform them some 'test' operations like subtraction or division. If we subtract the first quantity from the second quantity, the difference can be positive if the second is greater, negative if the first is greater, or zero if they are equal.

A similar situation comes if we divide the last by the first. If the ratio (division) is greater than one, then the last quantity is greater than the first if it's less than one, the first is greater, and if it's exactly one, they are equal.

Zoe caught 9 catfish and 6 trout on her first day. It produces a ratio of 9/6=1.5 catfish to trout relation. The second day she caught 17 catfish and 13 trout, which gives us a ratio of 17/13=1.3 catfish to trout.

Based on the ratios we can say Zoe caught a lower ratio of catfish to trout on the second day of her fishing trip

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V+v=gt
2v= g x 8
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3 years ago
200-grams of computer chips with a specific heat of 0.3 kJ/kg·K are initially at 25°C. These chips are cooled by placement in 0.
balu736 [363]

Answer:

a. -0.01324 kJ/K,  b.  = 0.03233 kJ/K , c.  = 0.01909, Yes the process is possible

Explanation:

Heat transfer will occur between the chip and the surrounding fluid. Then, finally they will attain a common equilibrium temperature and heat transfer will stop. Now, if we assume that, after heat transfer, chip will attain the temperature of fluid, that is, -34 C,, So , to check whether this is possible

Amount of energy lost by the chip = m . c . (T(i) - T(f))

= 0.2 x 0.3 (25 + 34) = 3.54 KJ

Now, to evaluate the final state of the fluid, after the heat transfer completion,

Energy Gained = m(mew final – mew initial) = m[(μf+ x . μfg) - μf]

Note that heat transfer will change the internal energy of the fluid. Do not consider enthalpy change, as this is not a problem involving fluid flow in and out of the system

M[(μf+ x . μfg) - μf] = m(xμfg)

<u>Energy gained by the fluid will be equal to the energy lost by the chip (No energy loss to the surroundings)</u>

3.54 = 0.1 . X x 203.29

<u>x = 0.1741, which is the dryness fraction of fluid at the final state.</u>

Observe that the total energy lost by the chips is 3.45 kJ and fluid R-134a has got its value of mew fg at -34 C which is = 203.29 kJ/kg

So for 0.1kg of R-134a

0.1 x μfg= <u>20.329 kJ, which is much greater than 3.45 kJ</u>, therefore, it is certain that the state of fluid will be at -34 C only and at the saturation pressure of 69.56 KPa. So the chip will come to attain the temperature of -34 C.  

a. Write the equation for the change of entropy in the chips

ΔSchips = mchips . c . ln(T2/T1), where mc is the mass of chips, c is the specific heat of chips, T2 is the temperature at state 2 and T1 is the temperature at state 1

Substitute mc = 0.2 kg, c = 0.3kJ/kg.K, T1 = 25 + 273, T2 = -34 + 273

delSchips = 0.2 x 0.3 x ln [(-34+273)/ (25+273)]

= -0.01324 kJ/K

There fore the change in entropy of the chips is -0.01324 kJ/K

b. Entropy change of fluid R- 134a

ΔS2 = m[Sfinal – S initial]

= m[Sf + x . Sfg - Sf]

= 0.2 x (0.1741 x 0.92859)

= 0.03233 kJ/K

c. Calculate the total change in the entropy of the entire system

delS = delSchips + delSR -134a

= -0.01324 + 0.03233

= 0.01909

<u>Since the total change in entropy of the entire system is positive that exactly explains that the actual processes are happening in the direction of increase of entropy therefore, the process is possible.</u>

<u />

6 0
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gtnhenbr [62]

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Explanation:

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Answer:

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<em>p</em> = 6.30 N

<em>f</em> = <em>p</em> = 6.30 N

5 0
3 years ago
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