Answer:
polar orbit is one in which a satellite passes above or nearly above both poles of the body being orbited (usually a planet such as the Earth, but possibly another body such as the Moon or Sun) on each revolution. It has an inclination of about 60 - 90 degrees to the body's equator.[1] A satellite in a polar orbit will pass over the equator at a different longitude on each of its orbits.
Launching satellites into polar orbit requires a larger launch vehicle to launch a given payload to a given altitude than for a near-equatorial orbit at the same altitude, due to the fact that much less of the Earth's rotational velocity can be taken advantage of to achieve orbit. Depending on the location of the launch site and the inclination of the polar orbit, the launch vehicle may lose up to 460 m/s of Delta-v, approximately 5% of the Delta-v required to attain Low Earth orbit. Polar orbits are a subtype of Low Earth orbits with altitudes between 200 and 1,000 kilometers.[1]
Explanation:
Answer:
I don't have one sorry
Explanation:
it's because I can't see the question I don't know it I'm sorry dude
In order to solve the total pressure that is exerted by the gases, we need to use the Dalton's Law of Partial pressures. These are the calculations that you need to find out the total amount of pressure exerted to the gases:
3.00atm (N2) + 1.80atm (O2) + 0.29atm (Ar) + 0.18atm (He) + 0.10atm (H),
add up all of that, and the answer would turn out to be: 5.37atm.
Answer: The correct answer is D. 273 Kelvin, 0 degrees Celsius, 32 degrees Fahrenheit.
Explanation:
Conversion of degree Celsius to Kelvin :
K=^oC+273
Conversion of degree Celsius to degrees Fahrenheit :
^oF=(\frac{9}{5}\times ^oC)+32
By using these two conversion factors, we get the three temperature readings all mean the same thing.
For option A :
K=^oC+273=100+273=373K
^oF=(\frac{9}{5}\times ^oC)+32=(\frac{9}{5}\times 100)+32=212^oF
For option B :
K=^oC+273=100+273=373K
^oF=(\frac{9}{5}\times ^oC)+32=(\frac{9}{5}\times 100)+32=212^oF
For option C :
K=^oC+273=0+273=273K
^oF=(\frac{9}{5}\times ^oC)+32=(\frac{9}{5}\times 0)+32=32^oF
For option D :
K=^oC+273=0+273=273K
^oF=(\frac{9}{5}\times ^oC)+32=(\frac{9}{5}\times 0)+32=32^oF
From the given options, only option (D) is correct.
Hence, the correct option is, (D) 273 Kelvin, 0 degrees Celsius, 32 degrees Fahrenheit
Hope this helps!
