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:
(2) Organelles must work together and their
activities must be coordinated
Explanation:
Organelles are usually located in cells. They are saddled with the role of performing specific functions in the cells for the overall functioning of life. In eukaryotic cells, the organelles are membrane bounded but in prokaryotic or primitive cells such is not the case.
Examples of cell organelles are ribosome, food vacuole, nucleus e.t.c. Just like organs in the body, organelles must work together in order to enhance life.
Since medals form cations
nonmedals form anions
Answer:
8.625 grams of a 150 g sample of Thorium-234 would be left after 120.5 days
Explanation:
The nuclear half life represents the time taken for the initial amount of sample to reduce into half of its mass.
We have given that the half life of thorium-234 is 24.1 days. Then it takes 24.1 days for a Thorium-234 sample to reduced to half of its initial amount.
Initial amount of Thorium-234 available as per the question is 150 grams
So now we start with 150 grams of Thorium-234
So after 120.5 days the amount of sample that remains is 8.625g
In simpler way , we can use the below formula to find the sample left
Where
is the initial sample amount
n = the number of half-lives that pass in a given period of time.
