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slava [35]
3 years ago
12

The spectrum of a distant star shows that one in 2 e6 of the atoms of a particular element is in its first excited state 7.5 eV

above the ground state. What is the temperature of the star? (You can ignore the other excited states and assume the ratio of statistical weights is 4
Physics
1 answer:
Alchen [17]3 years ago
5 0

Answer:

The temperature of star is 5473.87 K

Explanation:

Given:

Energy difference \Delta E = 7.5 eV

The ratio of number of particle \frac{N_{f} }{N_{i} } = \frac{1}{2 \times 10^{6} }

Degeneracy ratio \frac{g_{f} }{g_{i} }  = 4

From the formula of boltzmann distribution for population levels,

     \frac{N_{f} }{N_{i} } =\frac{g_{f} }{g_{i} }  e^{-\frac{\Delta E}{kT} }

Where k = boltzmann constant = 8.62 \times 10^{-5} \frac{eV}{K}

     \frac{1}{2 \times 10^{6} } =4  e^{-\frac{7.5 eV}{8.62 \times 10^{-5} T} }

  8 \times 10^{6} } = e^{\frac{7.5 eV}{8.62 \times 10^{-5} T} }

\ln(8 \times 10^{6})  = {\frac{7.5 eV}{8.62 \times 10^{-5} T} }

  T  = {\frac{7.5 eV}{8.62 \times 10^{-5} \ln(8 \times 10^{6})} }

  T = 5473.87 K

Therefore, the temperature of star is 5473.87 K

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4. A spring is stretched 0.5 m from equilibrium. The force constant (k) of the
makkiz [27]

Answer:

31.25

Explanation:

The formula for the potential energy of a spring is \frac{1}{2}kx^2, where x is the distance in meters the spring is stretched. Plugging in the numbers that you are given, you get \frac{1}{2}\cdot 250\cdot 0.25=31.25. Hope this helps!

8 0
3 years ago
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As an astronaut goes out into space, her mass _______________ and her weight ____________________
mel-nik [20]
Her mass stays the same and her weight changes
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All of the following are good tips for controlling your emotions EXCEPT:
TEA [102]

Answer:

B

Explanation:

Building up emotions can be bad for your health and hurt you mentally all other options are good.

8 0
3 years ago
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I was driving along at 20 m/s , trying to change a CD and not watching where I was going. When I looked up, I found myself 46 m
vampirchik [111]

Answer:

6.46393559312 m/s²

Explanation:

Time taken to cover 56 m

t=\dfrac{56}{29}\ s

Distance covered in 0.42 seconds

0.42\times 20=8.4\ m

From equation of linear motion

s=ut+\frac{1}{2}at^2

8.4+20(\dfrac{56}{29}-0.42)+\dfrac{1}{2}a(\dfrac{56}{29}-0.42)^2=46\\\Rightarrow a=(46-8.4-20(\dfrac{56}{29}-0.42))\times\dfrac{2}{(\dfrac{56}{29}-0.42)^2}\\\Rightarrow a=6.46393559312\ m/s^2

The minimum acceleration is 6.46393559312 m/s²

5 0
3 years ago
A block with mass m =6.4 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.28 m.
Zanzabum

Answer

given,

mass of block (m)= 6.4 Kg

spring is stretched to distance, x = 0.28 m

initial velocity = 5.1 m/s

a) computing weight of spring

    k x = m g

k = \dfrac{mg}{x}

k = \dfrac{6.4 \times 9.8}{0.28}

      k = 224 N/m

b) f = \dfrac{\omega}{2\pi}

    \omega = \sqrt{\dfrac{k}{m}}= \sqrt{\dfrac{224}{6.4}} = 5.92 \ rad/s

   f = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}

   f = \dfrac{1}{2\pi}\sqrt{\dfrac{224}{6.4}}

  f =0.94\ Hz

c)  v_b = -v cos \omega t

    v_b = -5.1 \times cos (5.92 \times 0.42)

    v_b = 4.04\ m/s

d)  a_{max} = v \omega

    a_{max} = 4.04 \times 5.92

    a_{max} =23.94\ m/s^2

e)  Y =- A sin (\omega t)

    A = \dfrac{v}{\omega}

    A = \dfrac{4.04}{5.92}

        A = 0.682 m

    Y =- 0.682 \times sin (5.92 \times 0.42)

    Y =- 0.42

Force =m \omega^2 |Y|

          =6.4 \times 5.92^2\times 0.42

F = 94.20 N

4 0
3 years ago
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