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3 years ago

Which is NOT an adaptation made by mammals during the Cenozoic era?

Physics

2 answers:

Anettt [7]3 years ago

3 0

Becoming Cold Blooded im pretty sure

Pachacha [2.7K]3 years ago

3 0

<h3> <u>Answer;</u></h3>

C) Becoming cold-blooded

<h3><u>Explanation;</u></h3>

<em><u>The Cenozoic era is an era that encompasses the past 65 million years of Earth history. </u></em>
<em><u>During this era mammals replaced reptiles as the dominant land animals. It is often called the age of mammals because land mammals came to dominate land life. Additionally, the development of flowering plants during this period influenced the evolution of birds and mammals.</u></em>
<em><u>The mammals were warm-blooded, and they could survive in cold regions and search for food during any season or time of day, they also adapted insulating body hair and more efficient lungs and hearts.</u></em>

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3 years ago

INK]: A car is moving at 200 km/hr east. What is its velocity?

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3 years ago

Assume that block A which has a mass of 30 kg is being pushed to the left with a force of 75 N along a frictionless surface. Wha

Veronika [31]

Answer:

The force of friction acting on block B is approximately 26.7N. Note: this result does not match any value from your multiple choice list. Please see comment at the end of this answer.

Explanation:

The acting force F=75N pushes block A into acceleration to the left. Through a kinetic friction force, block B also accelerates to the left, however, the maximum of the friction force (which is unknown) makes block B accelerate by 0.5 m/s^2 slower than the block A, hence appearing it to accelerate with 0.5 m/s^2 to the right relative to the block A.

To solve this problem, start with setting up the net force equations for both block A and B:

$F_{Anet} = m_A\cdot a_A = F - F_{fr}\\F_{Bnet} = m_B\cdot a_B = F_{fr}$

where forces acting to the left are positive and those acting to the right are negative. The friction force F_fr in the first equation is due to A acting on B and in the second equation due to B acting on A. They are opposite in direction but have the same magnitude (Newton's third law). We also know that B accelerates 0.5 slower than A:

$a_B = a_A-0.5 \frac{m}{s^2}$

Now we can solve the system of 3 equations for a_A, a_B and finally for F_fr:

$30kg\cdot a_A = 75N - F_{fr}\\24kg\cdot a_B = F_{fr}\\a_B= a_A-0.5 \frac{m}{s^2}\\\implies \\a_A=\frac{87}{54}\frac{m}{s^2},\,\,\,a_B=\frac{10}{9}\frac{m}{s^2}\\F_{fr} = 24kg \cdot \frac{10}{9}\frac{m}{s^2}=\frac{80}{3}kg\frac{m}{s^2}\approx 26.7N$

The force of friction acting on block B is approximately 26.7N.

This answer has been verified by multiple people and is correct for the provided values in your question. I recommend double-checking the text of your question for any typos and letting us know in the comments section.

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3 years ago

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Anna007 [38]

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Explanation:

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