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skelet666 [1.2K]
3 years ago
10

A chemical company claims that they are able to produce a certain product through a thermodynamically unfavored process. explain

how this is possible
Business
1 answer:
Alekssandra [29.7K]3 years ago
4 0
<span>It is definitely possible for a chemical company to produce a certain product through a thermodynamically unfavored process. An example of this process is endothermic reaction. The word endothermic means “taking in heat.” A constant input of energy, often in the form of heat, will keep an endothermic reaction going. For instance, an instant cold ice pack can get cold without putting it in the fridge or freezer. How is that? Endothermic chemical reaction.</span>
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The population p, in millions, of a certain country can be calculated by p=0.83t54, where t is the time in years, and t = 0 repr
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P(t) = (0.83)t^(5/4) 
<span>2096 - 2015 = 81 </span>
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<span>Population in 2096 expected to be 202 million.</span>
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According to John Firman, the research director for the International Association of Chiefs of Police (IACP), the most common re
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3 years ago
Derozan Corp. manufactured equipment at a cost of $366,953 and leased it to B Corp. on January 1, 2019 for an eight-year period
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Answer: $‭726,957.6‬0

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3 years ago
A body in the solar system has a period of 10,759.22 days and a perihelion speed of 10.18 km/s. a. Calculate the aphelion radius
anastassius [24]

Answer:

Explanation:

From the information given, by applying Kepler's 3rd law,

T^2 \alpha  a^3

where;

T = period

a = semi major axis

T = 356 days (for earth)

a = 1 AU = 1.496 \times 10^8 \ km

Therefore, T^2 = ca^3

c= \dfrac{365^2}{(1.496 \times 10^8)^3}

c = 3.9791 \times 10^{20} \ day^2/km^3

However, if the body in the solar system has a period of 10.759.22 days, then, a =?

∴

T^2 = ca^3

a3 = \dfrac{10759.22^2}{3.9791 \times 10^{-20}}

a^3 = 2.9092 \times 10^{27}

a= \sqrt[3]{2.9092 \times 10^{27}}

a = 1.4275 \times 10^9 \ km

However, the velocity for a perihelion = 10.18 km/s

Using the formula

v = \sqrt{GM ( \dfrac{2}{r}-\dfrac{1}{a})} to calculate the radius, we have:

G = 6.674 \times 10^{-11}

M = 1.989\times 10^{30} \ kg

r = perihelion

v ^2= GM ( \dfrac{2}{r}-\dfrac{1}{a})

(10.18 \times 10^3) ^2= 6.674 \times 10^{-11} \times 1.989 \times 10^{30}  ( \dfrac{2}{r}-\dfrac{1}{1.425 \times 10^{12}})

7.8068 \times 10^{-13}= \dfrac{2}{r}-\dfrac{1}{1.425 \times 10^{12}}

\dfrac{2}{r} = 1.4824 \times 10^{-12}

r = \dfrac{2}{1.4824 \times 10^{-12}}

r = 1.349 \times 10^{12}

Similarly, the perihelion is expressed by the equation,

r = a(1 - e)

where ;

e= eccentricity

∴

1.349 \times 10^{12} = 1.425 \times 10^{12} ( 1 - e)

1.349 \times 10^{12}  -  1.425 \times 10^{12}= -  1.425 \times 10^{12} (e)

-7.6\times 10^{10}= -  1.425 \times 10^{12} (e)

\dfrac{-7.6\times 10^{10}}{-  1.425 \times 10^{12}}=  (e)

e ( eccentricity) = 0.0533

Aphelion radius in natural miles, r = a( 1+ e)

r = 1.425 \times 10^{12} ( 1 + 0.0533)

r = 1.50 \times 10^{12} \ m

to nautical miles, we have:

r = 1.50 \times 10^{12} \times 0.00054  \ nautical \ mile

radius of aphelion \mathbf{r = 8.10 \times 10^8} nautical miles

In respect to the value of a( i.e 1.4275 \times  10^9 \ km)

the body of the solar system is Saturn

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