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denpristay [2]
3 years ago
15

Find the speed for a star in which this line appears at wavelength 120.2 nm . Express your answer to three significant figures a

nd include the appropriate units. v
Physics
1 answer:
levacccp [35]3 years ago
8 0

Answer:

The speed of the star is -3.45x10^{6}m/s

Explanation:

Spectral lines will be shifted to the blue part of the spectrum if the source of the observed light is moving toward the observer, or to the red part of the spectrum when it is moving away from the observer (that is known as the Doppler effect).

Redshift: \lambda_{measured}  >  \lambda_{0}

Blueshift: \lambda_{measured}  <  \lambda_{0}

The wavelength at rest is 121.6 nm (\lambda_{0} = 121.6nm)

Due to that shift the velocity of the star can be determined by means of Doppler velocity.

v = c\frac{\Delta \lambda}{\lambda_{0}} (1)

Where \Delta \lambda is the wavelength shift, \lambda_{0} is the wavelength at rest, v is the velocity of the source and c is the speed of light.

v = c(\frac{\lambda_{measured}- \lambda_{0}}{\lambda_{0}})

In this case \lambda_{measured} is equal to 120.2 nm.

v = (3x10^8m/s)(\frac{120.2 nm - 121.6 nm}{121.6 nm)})

v = -3.45x10^{6}m/s

Hence, the speed of the star is -3.45x10^{6}m/s.

Notice that the negative velocity means that the star is moving toward the observer.

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If one of the masses is doubled, the force of gravity between the objects is doubled. increases, the force of gravity decreases. If the distance is doubled, the force of gravity is one-fourth as strong as before.

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13. Convert 30° C to Fahrenheit<br> doo<br> a. 30 F<br> b. 86 F<br> 34 F<br> d. 100 F
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Answer:86F

Explanation:

Using the formula

F=(°Cx9/5)+32

F=(30x9/5)+32

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F=54+32

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6 0
3 years ago
Two 2.0cm×2.0cm metal electrodes are spaced 1.0 mm apart and connected by wires to the terminals of a 9.0 v battery. What is the
Fantom [35]

Answer:

The charge on each electrode is

q=0.354 x10^{-12}

Explanation:

The charge to find the equation is development

V=9v

d=1.0 mm\\A=2cm*2cm=0.02m*0.02m\\A=4x10^{-4} \\E_{o} =8.85x10^{-12} (\frac{C^{2} }{N*m^{2} } )

q=C*V

C=\frac{E_{o}*A }{d} \\C=\frac{8.85x10^{-12} *4x10^{-4}  }{0.01}=0.354  \\

q=0.354*9v\\q=3.186x10^{-12} C

3 0
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A wire is run through a large piece of paper, forming a perpendicular angle with the paper. A bunch of magentically senstive iro
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Answer:

Option C

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