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wlad13 [49]
2 years ago
11

The sun radiates most strongly at a wavelength of about 550 nm. a star that radiated most strongly at 400 nm would be

Physics
1 answer:
OLEGan [10]2 years ago
5 0
550 nm is in the visible spectrum. 400 nm is in the ultra-violet spectrum.  The shorter the wavelength, the bigger the energy of the light photon emitted. Therefore a star that emits 400 nm photons, is hotter than the Sun. A star that is hotter is either bigger than the Sun (and has about the same age), either younger than the Sun (and has about the same dimension). 
Therefore a star that radiates at 400 nm would be hotter, bigger and younger than the Sun.  
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A uniformly accelerated car passes three equally spaced traffic signs. The signs are separated by a distance d = 25 m. The car p
DedPeter [7]

Answer:

a) v_{1}=\frac{x_{2}-x_{1}  }{t_{2}-t_{1}  }=\frac{(2(\frac{25}{3})-\frac{25}{3} )m}{3.9s-1.3s}  =3.2051 \frac{m}{s}

b) v_{2}=\frac{x_{3}-x_{2}  }{t_{3}-t_{2}  }=\frac{(25-2(\frac{25}{3}) )m}{5.5s-3.9s}  =5.2083 \frac{m}{s}

c) a=\frac{v_{2}-v_{1}  }{t_{2}-t_{1}  } =\frac{5.2083m/s-3.2051m/s}{5.5s-3.9s} =1.252 \frac{m}{s^{2} }

Explanation:

<em><u>The knowable variables are </u></em>

d_{t}=25m

t_{1}=1.3 s

t_{2}=3.9 s

t_{3}=5.5 s

Since the three traffic signs are <u>equally spaced</u>, the <u>distance between each sign is \frac{25}{3} m</u>

a) v_{1}=\frac{x_{2}-x_{1}  }{t_{2}-t_{1}  }=\frac{(2(\frac{25}{3})-\frac{25}{3} )m}{3.9s-1.3s}  =3.2051 \frac{m}{s}

b) v_{2}=\frac{x_{3}-x_{2}  }{t_{3}-t_{2}  }=\frac{(25-2(\frac{25}{3}) )m}{5.5s-3.9s}  =5.2083 \frac{m}{s}

Since we know the velocity in two points and the time the car takes to pass the traffic signs

c) a=\frac{v_{2}-v_{1}  }{t_{2}-t_{1}  } =\frac{5.2083m/s-3.2051m/s}{5.5s-3.9s} =1.252 \frac{m}{s^{2} }

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