The answer to this question is Vitreous Body. The Vitreous
Body is a jelly form substance that fills the space on the eyeball of humans.
The vitreous body also helps transmit light in the retina and this also
functions to help maintain the shape of the eyeballs and make it stay in place.
Answer:
The three features of proteins are most important in determining their electrophoretic mobility is that the molecular size, net charge, and structural configuration have to be determined.
Explanation:
Hope this helps you out!
Answer:
1.98% or 2%
Explanation:
In US, 1 out of 10000 people has the disease. So, the frequency of the recessive alleles is 1/10000 = 0.0001
This term is q^2, so now we need to find the q term alone. q= = 0.01
Now, as p+q=1 in HW equations, then p=1-q, in this case p=1-0.01=0.99
So, the next part is determine what percentage is the heterozygous carriers and the heterozigous are in the equation named as 2pq
2pq=> 2(0.01)*(0.99)=0.0198 If we convert it to percentage is 0.0198 * 100= 1.98% aproximate to 2%
Answer:
Autotrophs, producers and Plants
Explanation:
Trees occupy a special position in the food web of most of all organisms. They can belong to several specie of plants and can vary in physical traits like height, fruit kind, fruiting non fruiting aspect etc. But three things that can describe trees are:
1: Autotrophs: This is because trees can prepare their own food through the process of photosynthesis and several other organisms depend upon them for food.
2: Producers: Being able to synthesize their own food, trees are known as producers because they not only prepare their own food but also food for many other organisms like humans.
3: Plants: Because trees are just a sub category or kind of plants, the terms plants can be used to describe several characteristics of trees.
Hope it helps!
No list to choose from. But maybe this will help
How To Use Kepler’s Third Law
The following is an explanation to help you figure out how to use your calculator to solve problems involving Kepler’s Third Law, and three practice problems to check yourself with.
• If I give you the planet’s orbital period:
– Type in the orbital period
– Hit the x2 key on your calculator OR
– Type in “^” followed by “2” and hit the Enter key.
– Next, type in “^” followed by “0.333” and hit the Enter key, OR
– Next, hit the “2nd Function Key” (often says “SHIFT” or “2nd Func” or “INV”) – Hit the xy key and type in 3. Then hit enter.
– The number you see is the planet’s average distance from the sun, in AU.
• IfIgiveyoutheplanet’saverageorbitaldistancefromthesun:
– Type in the orbital distance.
– Hit the xy key on your calculator, type in 3, and hit “Equals” or “Enter” OR – Type in “^” followed by “3” and hit the Enter key.
– If you see a “square root” key on your calculator hit it OR
– Type in “^” followed by “0.5” and hit the Enter key.
– The number you see is the planet’s orbital period, in years.
Examples
• UsetheseexamplestodetermineifyouareusingKepler’sThird Law correctly:
– An asteroid orbits the sun at a distance of 2.7 AU. What is its orbital period?
• Using a = 2.7 AU, you should get P = 4.44 years.
– A dwarf planet discovered out beyond the orbit of Pluto is known to have an
orbital period of 619.36 years. What is its average distance from the sun? • Using P = 619.36 years, you should get a = 72.66 AU.
– Chiron is a dwarf planet that orbits the sun between Saturn and Uranus, and has an average distance from the sun of 14 AU. What is its orbital period?
• Using a = 14.0 AU, you should get P = 52.38 years.