Answer:
I dont know and just write it for loginign brainly
Given data:
● Angular velocity of the turntable ω = 33 rev/min
Therefore,
ω = 33 rev/min
ω = 33rev/min × 2π rad/rev × 1 min/60 sec = 3.45 rad/s
● The distance of the watermelon seed from the axis of rotation r = 7.8 cm = 0.078 m
● μs = the coefficient of static friction
Section a:
The seed is undergoes circular motion and it is been effected by centripetal acceleration.
ac = rω^2
ac = 0.078 × 3.45^2
ac = 0.9284 m/s^2
Therefore,
the centripetal acceleration of the seed is 9.274 m/s^2
Section b:
If the seed is observed not to slip at the course of the circular motion, then the supplied frictional force given by the seed and surface of turntable would at least be equivalent to the centripetal force working on the seed.
Centripetal Force = Frictional Force
mrω^2 = μsmg
μs = rω^2 /g
μs = 0.078 × 3.45^2
------------
9.81
μs = 0.09464
Thus,
the coefficient of static friction is 0.09464.
Answer: The RNA strand is AUAAUCGGUCGCG
Explanation: The nucleotide sequence of RNA has four bases namely: Adenine A, Uracil U, Guanine G and Cytosine C. DNA nucleotide sequence has four bases namely: Adenine A, Thymine T, Cytosine C and Guanine G. In RNA, A pairs with U, and C pairs with G. In DNA, A pairs with T and C pairs with G.
Answer:
The Parents may have had heterozygous genotypes.
Explanation:
Normal wings could be the dominant allele that is expressed in the phenotype, and altered wings could be the recessive allele. If both parents had a heterozygous genotype, it is possible for the offspring to have a homzygous recessive genotype.
