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Dvinal [7]
3 years ago
12

An astronomer whose secret hobby is riding merry-go-rounds has dedicated his career to finding the stars that rotate the most ra

pidly. But the stars are all v_________________.
Physics
1 answer:
sp2606 [1]3 years ago
4 0

An astronomer whose secret hobby is riding merry-go-rounds has dedicated his career to finding the stars that rotate the most rapidly. But the stars are all very far and hence he can not identify the starts that spins more rapidly.

<u>Explanation:</u>

The things in the world rotates. the earth, sun star everything will be spinning. There is a star named Achenar. this is the start that spins at a faster rate. It is the biggest star that stays at the position 10. Its mass is 7 times greater than the sun's mass.The shining of the star can be measured by luminosity. It is the measure of the brightness of the star.

The energy that is given by a star in a second of time is the luminosity of the star. All stars are present in the different distances form the surface of the earth. but when we look from the earth or even when when go at certain altitude we can find all stars to be located at the same distance. hence if we need to find the star that spins more rapidly, then the star spectral lines can be observed. The stars that rotates will be having the wide lines in the spectra.

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A hill that has a 15.5% grade is one that rises 15.5 m vertically for every 100.0 m of distance in the horizontal direction. At
Andreas93 [3]

The angle of incline of the hill above the horizontal is 8.81°.

Since the hill has a 15.5% grade is one that rises 15.5 m vertically for every 100.0 m of distance in the horizontal direction.

<h3>Tangent of the angle of the incline of the hill,</h3>

The tangent of the angle of the incline of the hill, Ф is

tanФ = vertical rise/horizontal distance = grade of hill

Now, the vertical rise = 15.5 m and the horizontal distance = 100.0 m

So, substituting the values of the variables into the equation, we have

tanФ = vertical rise/horizontal run

tanФ = 15.5 m/100.0 m

tanФ = 0.155

<h3>Angle of incline of the hill</h3>

Taking inverse tan of both sides, we have

Ф = tan⁻¹(0.155)

Ф = 8.81°

So, the angle of incline of the hill above the horizontal is 8.81°.

Learn more about angle of incline of a hill here:

brainly.com/question/10056962

6 0
2 years ago
PLEASE HELP! It’s urgent... and please show your work!!
bagirrra123 [75]

Answer:

A) 3.8 x 10

Explanation:

4 0
3 years ago
A block of mass M is suspended from two identical springs of negligible mass, spring constant k, and unstretched length L. First
34kurt

Answer:

The answer is (C)X1=2X2

Explanation:

8 0
3 years ago
What is the kinetic energy of a 1130 kg truck that is moving with a velocity of 40 m/s?
Makovka662 [10]

Answer:

The answer is 904,000.

Kinetic energy=1/2mv^2.

1/2×1130×40^2.

1/2×1808000=904,000Joules.

8 0
3 years ago
Ultraviolet light of wavelength 270 nm strikes a metal whose work function is 2.3 eV.What is the shortest de Broglie wavelength
soldier1979 [14.2K]

Answer:

The shortest de Broglie wavelength for the electrons that are produced as photoelectrons is 0.81 nm

Explanation:

Given;

wavelength of ultraviolet light, λ = 270 nm

work function of the metal, φ = 2.3 eV = 2.3 x 1.602 x 10⁻¹⁹ J = 3.685 x 10⁻¹⁹ J

The energy of the ultraviolet light is given by;

E = \frac{hc}{\lambda}\\\\E = \frac{(6.626*10^{-34} )(3*10^{8}) }{270*10^{-9} }\\\\E = 7.362 * 10^{-19} \ J

The energy of the incident light is related to kinetic energy of the electron and work function of the metal by the following equation;

E = φ  + K.E

K.E = E - φ

K.E = (7.362 x 10⁻¹⁹ J) - (3.685 x 10⁻¹⁹ J )

K.E = 3.677 x 10⁻¹⁹ J

K.E = ¹/₂mv²

mv² = 2K.E

velocity of the electron is given by;

V = \sqrt{\frac{2K.E}{m} }\\\\V =  \sqrt{\frac{2(3.677*10^{-19}) }{9.1*10^{-31} } }\\\\V = 8.99*10^{5}  \ m/s

the shortest de Broglie wavelength for the electrons is given by;

\lambda = \frac{h}{mv}\\ \\\lambda = \frac{6.626*10^{-34} }{(9.1*10^{-31})( 8.99*10^{5} )}\\\\\lambda = 8.10*10^{-10} \ m\\\\\lambda = 0.81 \ nm

Therefore, the shortest de Broglie wavelength for the electrons that are produced as photoelectrons is 0.81 nm

7 0
3 years ago
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