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Alenkinab [10]
3 years ago
7

Which of the following cannot be determined from the spectrum of a star?

Physics
2 answers:
lozanna [386]3 years ago
7 0

Answer:

The chemicals in the core of the star

Explanation:

The spectrum of a star is of its atmosphere. The absorption or emission spectrum that is obtained indicates the chemical composition of the star's atmosphere. The peak wavelength obtained indicates the surface temperature of the star (using Wien's-Displacement Law). The shift in the spectrum indicates the motion of the star. If there is blueshift, it indicates that it is moving towards us where as if there is a red shift, it indicates that it is moving away from us.

grigory [225]3 years ago
4 0
The event that cannot be determined from the spectrum of a star is that the <span>chemicals in the atmosphere of the star</span>
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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.

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