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Snezhnost [94]
3 years ago
15

what are the two factors that create the difference between the temperate forest biome in the taiga biome

Biology
1 answer:
Vesna [10]3 years ago
8 0

The temperate forest biome covers latitudes ranging approximately from the southern United States to southern Canada, while the taiga biome, also known as boreal forest, extends from the latitude of southern Canada to about 60 degrees north latitude. (see References 1, References 3) Thus, these two biomes are adjacent, which explains the many similarities between taiga and northern temperate forests. Both biomes have four distinct seasons, but the temperate forest climates cover a much wider range of temperatures and precipitation patterns. Taiga, in contrast, is reliably cold: most of the precipitation falls as snow, winters are severe and the growing season is short -- about 130 days compared to 140 to 200 days for temperate forests.



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The sequences at the ends of linear chromosomes are called _______________. A protein that uses ATP hydrolysis to separate the t
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Answer: Telomeres, Helicases, Okazaki, DNA polymerase, Topoisomerase

Explanation:

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2. Helicases are proteins that uses energy (ATP) to unwind DNA strands during replication.

3. Okazaki fragments the small DNA nucleotide sequence synthesized separately on the lagging strand.

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Read 2 more answers
❥SCIENCE SUBJECT
Fiesta28 [93]

9514 1404 393

Answer:

  • time: 1.122 seconds
  • range: 10.693 m
  • maximum height: 1.543 m

Explanation:

<u>Given</u>:

  runner is launched at 30° angle to horizontal at 11 m/s

  acceleration due to gravity is g = -9.8 m/s²

<u>Find</u>:

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  runner's distance to the landing point

  runner's maximum height

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The (horizontal, vertical) speed components will be ...

  (11 m/s)(cos(30°), sin(30°)) = (5.5√3 m/s, 5.5 m/s)

The time of flight can be found from the height formula:

  h(t) = 1/2gt² +vt . . . . . . where v is the vertical speed at launch

The time we're concerned with is the time when h(t)=0 and t>0.

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The second factor is zero when ...

  t = (5.5√3)/4.9 ≈ 1.122 . . . seconds hang time

__

The distance to the landing point will be the product of horizontal speed and hang time:

  d = (5.5 m/s)(5.5√3/4.9 s) ≈ 10.693 m . . . . distance to landing

__

The maximum height can be found from the formula (based on conversion of kinetic energy to potential energy) ...

  h = v²/|2g| = (5.5 m/s)²/(2(9.8 m/s²)) ≈ 1.543 m . . . . maximum height

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a weightlifter lifts a 20 kilogram barbell above his head

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