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Angelina_Jolie [31]
3 years ago
9

A common radio wavelength observed coming from astronomical objects is 21 cm. What temperature is associated with this radiation

?
Chemistry
1 answer:
qaws [65]3 years ago
8 0

Answer:

The temperature associated with this radiation is 0.014K.

Explanation:

If we assume that the astronomical object behaves as a black body, the relation between its <em>wavelength</em> and <em>temperature</em> is given by Wien's displacement law.

\lambda_{max}=\frac{b}{T}

where,

λmax is the wavelength at the peak of emission

b is Wien's displacement constant (2.89×10⁻³ m⋅K)

T is the absolute temperature

For a wavelength of 21 cm,

T=\frac{b}{\lambda _{max} } = \frac{2.89 \times 10^{-3} m.K  }{0.21m} =0.014K

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b) velocity.

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e) increasing or decreasing

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The weight of the body decrease inside water why?​
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Penurunan atau kehilangan massa otot bisa menimbulkan penurunan berat badan yang tidak direncanakan

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3 years ago
Alex is on a special diet and needs to watch her intake of carbohydrates. Her doctor told her to maintain a 1,500 calorie per da
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Answer: Alex should get 300 Calories from carbohydrates, which is about 33 grams. This snack is almost 70% of her total grams of carbohydrates per day.

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3 years ago
The water in a pressure cooker boils at a temperature greater than 100°C because it is under pressure. At this higher temperatur
vazorg [7]

Answer:

the activation energy Ea = 179.176 kJ/mol

it will take  7.0245 mins for the same food to cook in an open pot of boiling water at an altitude of 10000 feet.

Explanation:

From the given information

T_1 = 100^0 C = 100+273 = 373 \ K \\ \\  T_2 = 113^0 C = 113 + 273 = 386 \ K

R_1 = \dfrac{1}{7}

R_2 = \dfrac{1}{49}

Thus; \dfrac{R_2}{R_1} = 7

Because at 113.0°C; the rate is 7 time higher than at 100°C

Hence:

In (7) = \dfrac{Ea}{8.314}( \dfrac{1}{373}- \dfrac{1}{386})

1.9459 = \dfrac{Ea}{8.314}* 9.0292  *10^{-5}

1.9459*8.314 = Ea * 9.0292*10^{-5}

16.1782126= Ea * 9.0292*10^{-5}

Ea = \dfrac{16.1782126}{ 9.0292*10^{-5}}

Ea = 179.176 kJ/mol

Thus; the activation energy Ea = 179.176 kJ/mol

b)

here;

T_2 = 386 \  K  \\ \\T_1 = (89.8 + 273)K = 362.8 \ K

In(\dfrac{R_2}{R_1})= \dfrac{Ea}{R}(\dfrac{1}{T_1}- \dfrac{1}{T_2})

In(\dfrac{R_2}{R_1})= \dfrac{179.176}{8.314}(\dfrac{1}{362.8}- \dfrac{1}{386})

In (\dfrac{R_2}{R_1}) = 0.00357

\dfrac{R_2}{R_1}= e^{0.00357}

\dfrac{R_2}{R_1}= 1.0035

where ;

R_2 = \dfrac{1}7{}

R_1 = \dfrac{1}{t}

Now;

\dfrac{t}{7}= 1.0035

t = 7.0245 mins

Therefore; it will take  7.0245 mins for the same food to cook in an open pot of boiling water at an altitude of 10000 feet.

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2.38×10^-3

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