A geostationary orbit can be achieved only at an altitude very close to 35,786 km (22,236 mi) and directly above the equator. This equates to an orbital velocity of 3.07 km/s (1.91 mi/s) and an orbital period of 1,436 minutes, which equates to almost exactly one sidereal day (23.934461223 hours).
That there are not significate amounts of dark matter wihtin our solar system.
- The dark matter is a component of the universe whose presence is discerned from its gravitational attraction rather than its luminosity.
- The force of attraction between all masses in the universe; especially the attraction of the earth's mass for bodies near its surface is known as gravitational attraction.
- The luminous normal matter, such as protons, neutrons, electrons, and atoms in the universe, there are about 4 grams of nonluminous normal matter, mainly intergalactic hydrogen and helium.
To know more about intergalactic
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Answer:
so angular velocity is 7.13128 sec−1
Explanation:
velocity v = 2.2 m/s
displacement s = 220 mm = 0.220 m
distance d = 510 mm = 0.510 m
to find out
angular velocity
solution
we know that
angular velocity will be velocity ( v) / (displacement² + distance²) .....1
now put all these value in equation 1 and we get angular velocity i.e.
angular velocity = velocity ( v) / (displacement² + distance²)
angular velocity = 2.2 / (0.22² + 0.51²)
angular velocity = 2.2 / 0.3085
angular velocity = 7.13128
so angular velocity is 7.13128 sec−1
A) We want to find the work function of the potassium. Apply this equation:
E = 1243/λ - Φ
E = energy of photoelectron, λ = incoming light wavelength, Φ = potassium work function
Given values:
E = 2.93eV, λ = 240nm
Plug in and solve for Φ:
2.93 = 1243/240 - Φ
Φ = 2.25eV
B) We want to find the threshold wavelength, i.e. find the wavelength such that the energy E of the photoelectrons is 0eV. Plug in E = 0eV and Φ = 2.25eV and solve for the threshold wavelength λ:
E = 1243/λ - Φ
0 = 1243/λ - Φ
0 = 1243/λ - 2.25
λ = 552nm
C) We want to find the frequency associated with the threshold wavelength. Apply this equation:
c = fλ
c = speed of light in a vacuum, f = frequency, λ = wavelength
Given values:
c = 3×10⁸m/s, λ = 5.52×10⁻⁷m
Plug in and solve for f:
3×10⁸ = f(5.52×10⁻⁷)
f = 5.43×10¹⁴Hz
Hertz, which is a unit for frequency (f), also means 1/second. Hence, the speed (S) of wave is the product of its wavelength (n) generated per second (frequency). The speed of the given wave above is,
S = nf ; S = (45 m) x (9 / s) = 405 m/s
Hence, the speed of the wave is 405 m/s.