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creativ13 [48]
3 years ago
14

Geologists have evidence that the continents were once a single giant landmass that land mass eventually split apart and the ind

ividual continents move to their current positions what role might this movement of continents have played in evolution?
Physics
1 answer:
mars1129 [50]3 years ago
7 0
The part that it played was in the changing of the environment/climate for the organisms that live on those continents. And the part it could've played was the way that the organisms had to adapt to that climate and it stayed that way over generations.

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How do I do this problem?
Aleks04 [339]

Use Charles Law: V1/T1 = V2/T2

0.30 m^3/27 C = V2/127 C

27V2 = 127 * 0.3

V2= 38.1/27 = 1.4 m^3

5 0
2 years ago
Which of the following statements correctly described natural electric fields?
Mila [183]

Thunder is evidence of electric fields in the atmosphere.

The charge difference between the clouds and earth can be described as parallel conducting plates that produce electric fields in the atmosphere.

could be true

5 0
3 years ago
In the mixtures lab, you waited to see if the liquid would form lumpy or fluffy masses, which would indicate it was a _____.
Over [174]
B) colloid because i took the test and that the answer

5 0
3 years ago
Read 2 more answers
In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at 78.5 m/s. Th
Andrews [41]

Answer:

FT is 1020.6 meters (1640.6 meters - 620 meters) far from MB

Explanation:

First you have to consider that the Ford Thunderbird (FT) follows a rectilinear motion with varying acceleration, while Mercedez Benz (MB) has a constant velocity (no acceleration). So if you finde the time spent by FT in each section, and the distance, then you will find the distance for MB.

1) Vf² = Vi² + 2ad, where Vf: final velocity, Vi: ionitial velocity, a: acceleration and d: distance.

For the first portion  (0 m/s)² = (78.5 m/s)² + 2a(250 m) ⇒

-(78.5 m/s)² / 2(250m) = a ⇒ a = -12.3 m/s².

Now, you can find the corresponding time for this section with the following formule: Vf = Vi + at ⇒ 0 m/s = 78.5 m/s + (-12.3 m/s²) t

⇒ t= (-78.5 m/s)/ (-12.3 m/s²) ⇒ t= 6.4 seconds.

2) Then FT spent 5 seconds in the pit.

3) The the FT accelerates until reach 78.5 m/s again in a distance of 370 m.

Vf² = Vi² + 2ad ⇒ (78.5 m/s)² = (0 m/s)² + 2a(370 m)

⇒ (78.5 m/s)²/ 2(370 m) = a ⇒ a = 8.3 m/s²

Then, Vf = Vi + at ⇒ 78.5 m/s = 0 m/2 + (8.3 m/s²) t

⇒ (78.5 m/s)/(8.3 m/s²) = t ⇒ t = 9.5 seconds.

4) Summarizing, the FT moves 620 meters (250 + 370 mts) in 20.9 seconds ( 6.4 s + 5 s + 9.5 s).

5) During this time, MB moves

Velocity = distance/ time ⇒ Velocity x time = Distance

⇒ Distance = (78.5 m/s) x  (20.9 seconds) ⇒ Distance = 1640.6 meters

6) Finally, the FT is 1020.6 meters (1640.6 meters - 620 meters) far from MB

3 0
3 years ago
Biologists have studied the running ability of the northern quoll, a marsupial indigenous to Australia. In one set of experiment
Drupady [299]

Answer:

u_K=0.862

Explanation:

The force of friction between the quails feet and the ground is:

F=m*a

F_K=m*a

F_K=u_k*m*g

u_K*m*g=m*a_c

u_K*g=a

u_K=\frac{a_c}{g}

a_c=\frac{v^2}{r}

So the coefficient of static is solve

u_K=\frac{\frac{v^2}{r}}{g}

u_K=\frac{v^2}{r*g}=\frac{(2.6m/s)^2}{0.80m*9.8m/s^2}

u_K=0.862

4 0
3 years ago
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