Pls Help semester ends at 11
2. A 2 kg bowling ball rolls at a speed of 5 m/s on the roof of the building that is 40 m tall.


Circle one: KE / GPE / both


Show your work for finding the values of each type of energy the object has:

Answers

Answer 1

Answer:  Both

Explanation:

KE can be solve using the formula:

ke = 0.5 mv^2

where m is the mass of the object

v is the velocity

ke = 0.5 ( 2 kg ) ( 5 m/s)^2

ke = 25 J

the GPE can be solved using the formula:

GPE = mgh

where m is the mass

g is the acceleration due to gravity ( 9.81 m/s^2)

h is the height

GPE = ( 4 kg)(9.81 m/s^2)( 40 m)

GPE = 392.4 J

I hope this helps, have a great day!!


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when a tautomeric shift occurs, what is true about the resulting mutation?

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When a tautomeric shift occurs, then the resulting mutation ALWAYS changes a purine to a purine or a pyrimidine to a pyrimidine. A tautomeric shift may lead to a point mutation.

A mutation is any alteration in the nucleotide (DNA) sequence of the genome of an individual.

A tautomeric shift can be defined as a temporary change in the nucleotide base structure.

A transition mutation is a point mutation of one purine or pyrimidine base pair to the other purine/pyrimidine that occurs when one nucleotide base undergoes a tautomeric shift during DNA replication.

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b
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d
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Answer: glucose I think?

Explanation: I’m not sure but it should be more efficient.

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A test cross performed between the individual with unkown genotype and the h0m0zyg0us recessive individual. 9) hh. 10) Hh or HH. 11) HH x hh. 12) 100% Hh. 13) Hh x  hh. 14) 50%Hh and 50% hh. 15) -16) in the text.

-----------------------------------

Available data:Single diallelic gene codes for hair in Guinea pigsAllele H is dominant and codes for having hairAllele h is recessive and codes for not having hair

For this gene we will assume complete dominance, meaning that the dominant alele completely hides the expression of the recessive allele.

According to this information, we can say that

Genotype      Phenotype

HH                  Hairy pig

Hh                   Hairy pig

hh                   Hairless pig

9) Geneviene does not have hair  ⇒  Its genotype is hh

10) Fred has hair ⇒ Its genotype is HH or Hh

Possible crosses

11 and 12)

Cross1: If Fred was genotype HH and bred with Genevieve

Parentals) HH    x    hh

Gametes)   H   H    h    h

Punnett square)    H    H

                        h   Hh   Hh

                        h   Hh   Hh

F1) 100% of the progeny is expected to have hair and be heter0zyg0us, Hh.

13 and 14)

Cross2: If Fred was genotype Hh and bred with Genevieve

Parentals) Hh    x    hh

Gametes)   H   hh    h    h

Punnett square)    H    h

                        h   Hh   hh

                        h   Hh   hh

F1) 50% of the progeny is expected to have hair and be heter0zyg0us, Hh.

     50% of the progeny is expected to be hairless and be h0m0zyg0us recessive, hh.

15)  

• If the individual with the unknown genotype (Fred) is heter0zyg0us, the phenotypic rate of the descendants is 50% heter0zyg0us -Hh- and 50% h0m0zyg0us recessive -hh-.

• But if the individual with the unknown genotype is h0m0zyg0us dominant, the phenotypic rate of the descendants is 100% heter0zyg0us for the trait, Hh.  

16)

The weakness is that even if Fred is heter0zyg0us for the trait, when making the cross it might occur that all individuals in the progeny are born h0m0zyg0us recessive, and there is a risk of missinterpreting the results.

So in that situation, many crosses must be done between fred and genevieve to se sure about Freds genotype.

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Answer:

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Explanation:

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Explanation:

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Answer:

[tex][tex]\purple{\rule{45pt}{7pt}}\purple{\rule{45pt}{999999pt}}[tex][/tex]

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