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tekilochka [14]
3 years ago
11

Help Meeee Please!!!!! This was due so long ago... DO NOT REPORT, THERE IS NO REASON TO REPORT! Assignment: Weathering in Action

Exploration
You can see science in action all around you. In this activity, you will observe weathering.

Signs of Weathering| Mechanical or Chemical? Description of location

Flakes of brick -----

Cracks in rocks -----

Lichen growing on rocks------

Roots or plant growth ---------
making cracks in concrete

Worn away statues-------

Exposed rocks ----------

Rust on rocks -----------



Example You Found:







Sketches

Type of weathering:


Sketch:










Type of weathering:


Sketch:










Type of weathering:


Sketch:








I really need help, I don't understand this so please help me.
Below is the document to this it is the exact same as the question.
Physics
1 answer:
zaharov [31]3 years ago
3 0

Answer: Chemical

Explanation: because it is changeing its chemical structure if its wrong sorry .

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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
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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.

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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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