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o-na [289]
3 years ago
14

The H line in Calcium is normally at 396.9 nm. However, in a star's spectrum, it is measured at 398.1nm. How fast is the star mo

ving and is it moving towards the Earth or away from the Earth?
1- 904.3 km/s away from the Earth
2- 904.3 km/s towards the Earth
3- 907.0 km/s towards the Earth
4- 907.0 km/s away from the Earth
Physics
1 answer:
agasfer [191]3 years ago
3 0

As we know by Doppler's effect of light we have

\frac{\Delta \lambda}{\lambda} = \frac{v}{c}

here we will have

[tex}\frac{398.1 nm - 396.9 nm}{398.1 nm} = \frac{v}{c}[/tex]

here by solving above we have

3.01 \times 10^{-3} = \frac{v}{c}

here we have

v = 904.3 km/s

since wavelength is increased so we can say that it is moving away

so correct answer is

1- 904.3 km/s away from the Earth

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Answer:

The height of Sears Tower is 1448.5 feet.

Explanation:

<h3>We apply the free fall formula to the ball: </h3><h3>y=v_{o} *t+\frac{1}{2} *g*t^{2}</h3><h3>y: The vertical distance the ball moves at time t  </h3><h3>v_{o}i: Initial speed </h3><h3>g=Gravity acceleration=9.8*(\frac{\frac{1ft}{0.305m} }{s^{2} } )</h3>

Known information

We know that the vertical distance (y) that the ball moves in 9,5s  is equal to height of Sears Tower (h).  

Too we know that the ball is released from rest, then,v_{0}=0

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We replace  in the equation 1 the data following;

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Answer:

\ \text{m/s}

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The collision is perfectly inelastic as the lumps stick to each other so we have the relation

mu_1+mu_2=(m+m)v\\\Rightarrow m(u_1+u_2)=2mv\\\Rightarrow v=\dfrac{u_1+u_2}{2}\\\Rightarrow v=\dfrac{3x+3y-3z-4x+0y-4z}{2}\\\Rightarrow v=-0.5x+1.5y-3.5z=\ \text{m/s}

The velocity of the stuck-together lump just after the collision is \ \text{m/s}.

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