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goblinko [34]
1 year ago
15

The spectrum of Star A has an absorption line of hydrogen at 660.0 nm. The spectrum of Star B has an absorption line at 666 nm.

The wavelength of this transition in the laboratory is 656.2 nm. What can I say about Star A and B?
Physics
1 answer:
rjkz [21]1 year ago
6 0

Answer:

The stars are moving away from us.

Explanation:

The observed wavelengths of hydrogen transition for stars A and B (660.0 nm and 666 nm respectively) are greater than that observed in the laboratory (656.2 nm). The observed long wavelengths for the stars means that the light from the stars is red-shifted.

According to the Doppler effect, red-shifted light means that the source is moving a way from the observer; therefore, we arrive at the conclusion that the stars A and B are moving away from us.

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Momentum = Mass x Velocity

Put the values where they belong and solve for Velocity.

In this case, since Mass is being multiplied by Velocity, to solve for be Velocity you would divide both sides by Velocity.  The velocity will equal the momentum divided by the mass.

4 0
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URGENT!!!!
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8 cm

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2 years ago
A 0.300kg glider is moving to the right on a frictionless, ­horizontal air track with a speed of 0.800m/s when it makes a head-o
e-lub [12.9K]

Answer:

The final velocity of the first glider is 0.27 m/s in the same direction as the first glider

The final velocity of the second glider is 1.07 m/s in the same direction as the first glider.

0.010935 J

0.0858675 J

Explanation:

m_1 = Mass of first glider = 0.3 kg

m_2 = Mass of second glider = 0.15 kg

u_1 = Initial Velocity of first glider = 0.8 m/s

u_2 = Initial Velocity of second glider = 0 m/s

v_1 = Final Velocity of first glider

v_2 = Final Velocity of second glider

As momentum and Energy is conserved

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}

{\tfrac {1}{2}}m_{1}u_{1}^{2}+{\tfrac {1}{2}}m_{2}u_{2}^{2}={\tfrac {1}{2}}m_{1}v_{1}^{2}+{\tfrac {1}{2}}m_{2}v_{2}^{2}

From the two equations we get

v_{1}=\frac{m_1-m_2}{m_1+m_2}u_{1}+\frac{2m_2}{m_1+m_2}u_2\\\Rightarrow v_1=\frac{0.3-0.15}{0.3+0.15}\times 0.8+\frac{2\times 0.15}{0.3+0.15}\times 0\\\Rightarrow v_1=0.27\ m/s

The final velocity of the first glider is 0.27 m/s in the same direction as the first glider

v_{2}=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 0.3}{0.3+0.15}\times 0.8+\frac{0.3-0.15}{0.3+0.15}\times 0\\\Rightarrow v_2=1.067\ m/s

The final velocity of the second glider is 1.07 m/s in the same direction as the first glider.

Kinetic energy is given by

K=\frac{1}{2}m_1v_1^2\\\Rightarrow K=\frac{1}{2}0.3\times 0.27^2\\\Rightarrow K=0.010935\ J

Final kinetic energy of first glider is 0.010935 J

K=\frac{1}{2}m_2v_2^2\\\Rightarrow K=\frac{1}{2}0.15\times 1.07^2\\\Rightarrow K=0.0858675\ J

Final kinetic energy of second glider is 0.0858675 J

6 0
2 years ago
No person who thinks scientifically places any faith in the predictions of astrologers. Nevertheless there are many people who r
Lilit [14]
Therefore, it can be reasonably concluded according to your
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2 years ago
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Which of the following sketches represents a possible configuration for this problem?
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Where are the following sketches?
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2 years ago
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