Answer:
<u>Starch</u> is the storage form of glucose (energy) in plants and the glucose molecules are linked by alpha 1,4 glycosidic linkage.
<u>Cellulose </u>is a structural component of the plant cell wall and glucose molecules are linked by beta 1,4 glycosidic linkage.
<u>Glycogen</u> is the storage form of glucose (energy) in animals and glucose molecules are linked by alpha 1,6 glycosidic linkage.
Explanation:
All of these sugars are polysaccaride sugars containing large number of glucose subunits.
Starch is a polysaccharide extracted from agricultural raw materials. It contains amylose and amylopectin. Amylose is an un-branched chain polymer of D-glucose units while amylopectin is a branched chain polymer of D-glucose units.
Glycogen is the storage form of glucose in animals, It is stored in muscles and liver and it is a branched polysaccaride.
Cellulose is the storage form of glucose in plants and leaves.
The dominance or recessivity associated with a particular allele is the result of masking, by which a dominant phenotype hides a recessive phenotype. By this logic, in heterozygous offspring only the dominant phenotype will be apparent.
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4.) We are told that ball A is travelling from right to left, which we will refer to as a positive direction, making the initial velocity of ball A, +3 m/s. If ball B is travelling in the opposite direction to A, it will be travelling at -3 m/s. The final velocity of A is +2 m/s. Using the elastic collision equation, which uses the conservation of linear momentum, we can solve for the final velocity of B.
MaVai + MbVbi = MaVaf + MbVbf
Ma = 10 kg and Mb = 5 kg are the masses of balls A and B.
Vai = +3 m/s and Vbi = -3 m/s are the initial velocities.
Vaf = +2 m/s and Vbf = ? are the final velocities.
(10)(3) + (5)(-3) = (10)(2) + 5Vbf
30 - 15 = 20 + 5Vbf
15 = 20 + 5Vbf
-5 = 5 Vbf
Vbf = -1 m/s
The final velocity of ball B is -1 m/s.
5.) We are now told that Ma = Mb, but Vai = 2Vbi
We can use another formula to look at this mathematically.
Vaf = [(Ma - Mb)/(Ma + Mb)]Vai + [(2Mb/(Ma + Mb)]Vbi
Since Ma = Mb we can simplify this formula.
Vaf = [(0)/2Ma]Vai + [2Ma/2Ma]Vbi
Vaf = Vbi
Vbf = [(2Ma/(Ma + Mb)]Vai + [(Ma - Mb)/(Ma + Mb)]Vbi
Vbf = [2Mb/2Mb]Vai + [(0)/2Mb]Vbi
Vbf = Vai
Vaf = Vbi
Vbf = 2Vbi
If the initial velocity of A is twice the initial velocity of B, then the final velocity of A will be equal to the initial velocity of B.
If the initial velocity of A is twice the initial velocity of B, then the final velocity of B will be twice the initial velocity of B.
