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11111nata11111 [884]
2 years ago
12

The international Space Station (ISS) orbits the Earth once every 90 mins at an altitude of 409 km. How high would it have to be

in order to be to be in geosynchronous orbit
Physics
1 answer:
Oksi-84 [34.3K]2 years ago
6 0

It would have to be 36,719 Km high in order to be to be in geosynchronous orbit.

To find the answer, we need to know about the third law of Kepler.

<h3>What's the Kepler's third law?</h3>
  • It states that the square of the time period of orbiting planet or satellite is directly proportional to the cube of the radius of the orbit.
  • Mathematically, T²∝a³
<h3>What's the radius of geosynchronous orbit, if the time period and altitude of ISS are 90 minutes and 409 km respectively?</h3>
  • The time period of geosynchronous orbit is 24 hours or 1440 minutes.
  • As the Earth's radius is 6371 Km, so radius of the ISS orbit= 6371km + 409 km = 6780km.
  • If T1 and T2 are time period of geosynchronous orbit and ISS orbit respectively, a1 and a2 are radius of geosynchronous orbit and ISS orbit, as per third law of Kepler, (T1/T2)² = (a1/a2)³
  • a1= (T1/T2)⅔×a2

           = (1440/90)⅔×6780

           = 43,090 km

  • Altitude of geosynchronous orbit = 43,090 - 6371= 36,719 km

Thus, we can conclude that the altitude of geosynchronous orbit is 36,719km.

Learn more about the Kepler's third law here:

brainly.com/question/16705471

#SPJ4

   

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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

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weeeeeb [17]

Answer:

1.) 4m

2.) 37 m

3.) 62m

4.) 2.5 s

Explanation:

1.) Given that the

Thinking distance = 1m

Breaking distance = 3m

Stopping distance = breaking distance + thinking distance

Stopping distance = 1 + 3 = 4m

2.) Given that the

Stopping distance = 52 m

Thinking distance = 15m

Breaking distance = 52 - 15 = 37m

3.) The stopping distance = 76m

Thinking distance = 14m

Breaking distance = 76 - 14 = 62m

It take the brakes 62m to slow the car down to a stop.

4.) Given that a lorry travels 28m when stopping from a speed of 4m/s. If its braking distance was 18m, what was the driver’s reaction time?

Thinking = stopping distance - braking distance

Thinking distance = 28 - 18 = 10m

Speed = distance/time

4 = 10/reaction time

Reaction time = 10/4

Reaction time = 2.5 s

5.) Question incomplete

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