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UkoKoshka [18]
3 years ago
10

Two animals that can be found at the Grand Canyon include the Abert’s and Kaibab squirrels. The Abert’s squirrel is found at the

south rim of the canyon and in a few surrounding states. The Kaibab squirrel, however, can only be found at the canyon’s north rim, mainly on the Kaibab Plateau. The Kaibab squirrel is not found anywhere else in the world. Scientists believe that the squirrels are actually closely related and have developed separately as a result of geographic isolation. It is possible that the squirrels were separated by plateau erosion that separated the forests and isolated the Kaibab population. Some scientists are studying if the squirrels are now two separate species or if the Kaibab squirrel is simply a subspecies of the Abert’s squirrel. Separate species are unable to reproduce with each other. Which evidence would be used to support the claim that the squirrels are two separate species? The Abert’s squirrel is found at the south rim of the canyon, but the Kaibab squirrel is found at the north rim of the canyon. Geologists prove that plateau erosion is the main cause of forest separation in the region. Scientists observe that Abert’s squirrels are slightly larger and have bushier tails than Kaibab squirrels do. When placed together in pairs for study, no pair of Abert’s squirrel and Kaibab squirrel result in offspring.
Physics
1 answer:
Tom [10]3 years ago
6 0
D. <span>When placed together in pairs for study, no pair of Abert's squirrel and Kaibab squirrel result in offspring.</span>
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Varvara68 [4.7K]

The question is incomplete. Here is the complete question:

A minivan is tested for acceleration and braking. In the street-start acceleration test, the elapsed time is 8.6 s for a velocity increase from 10 km/h to 100 km/h. In the braking test, the distance traveled is 44 m during braking to a stop from 100 km/h. Assume constant values of acceleration and deceleration. Determine  

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(b) the deceleration during the braking test.

Answer:

(a) 37500 km/h²

(b) 113636.36 km/h²

Explanation:

part (a)

Because it is given that we can assume constant acceleration therefore we can use the following equation of motion:

<em>v = u + (a)(t) </em>

where <em>v </em>is final velocity, <em>u </em>is initial velocity, <em>a </em>is acceleration and <em>t </em>is time change

Given in the question:

v = 100km/h

u = 10 km/h

t = 8.6 sec (changing to hours)

t = 0.0024 hours (round off to 4 decimal places)

100 = 10 + (a x 0.0024)

Rearranging the equation to find value of a

a = (100 – 10) / 0.0024

a = 37500 km/h² (Answer)

part (b)

Now we can use the following equation to find deceleration

<em>2(a)(s) = v² – u²</em>

Where a is acceleration, s is distance travelled, v is final velocity and u is initial velocity

Given in the question

s = 44 m

changing to km

s = 0.044 km

v = 0 km/h (because it stops)

u = 100 km/h

2(a)(0.044) = (0)² – (100)²

0.088(a) = 0 - 10000  

a = - 10000/0.088

a = - 113636.36 km/h2  

The negative sign in the answer shows that it is deceleration

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