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Leokris [45]
2 years ago
15

Pam's family is clearing some land, and there is a large boulder sitting on top of the ground. Even with the entire family pushi

ng on it, they cannot mov
it. Which type of friction exists between the boulder and the ground

Chemistry
1 answer:
SashulF [63]2 years ago
6 0

Answer:

d

Explanation:

I believe the answer to this question is d , I believe this as the question doesn't explain properly the surface under the boulder. it is questionable as the question is not clear though I believe this as there must be friction under the boulder stopping it from moving.

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Put the list in chronological order (1–5).
Leokris [45]

Explanation:

Filtration is a separation technique in which solid particles suspended in liquid medium are separated by allowing the mixture through the pores of the filter paper. By this solid particles get collect on filter paper and liquid drains out from the pores of the filter paper.

The chronological order for given steps will be:

  1. Weigh and fold the filter paper.
  2. Place the filter paper in the funnel, then place the funnel in the Erlenmeyer flask.
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Explain the significance of this quote, “Heisenberg may have slept here.”
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Na+H2O=NaOH+H2<br> Balancing
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Answer:

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3 years ago
30) Which of these statements are true?
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D) 1 and 4
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An alloy with an average grain diameter of 35 μm has a yield strength of 163 Mpa, and when it undergoes strain hardening, the gr
Lera25 [3.4K]

Answer:

\sigma_y\ =210.2\ MPa  

Explanation:

Given that

d= 35 μm ,yield strength = 163 MPa

d= 17 μm ,yield strength = 192 MPa

As we know that relationship between diameter and yield strength

\sigma_y=\sigma_o+\dfrac{K}{\sqrt d}

\sigma_y\ =Yield\ strength

d = diameter

K =Constant

\sigma_o\ =material\ constant

So now by putting the values

d= 35 μm ,yield strength = 163 MPa

163=\sigma_o+\dfrac{K}{\sqrt 35}      ------------1

d= 17 μm ,yield strength = 192 MPa

192=\sigma_o+\dfrac{K}{\sqrt 17}           ------------2

From equation 1 and 2

192-163=\dfrac{K}{\sqrt 17}-\dfrac{K}{\sqrt 35}

K=394.53

By putting the values of K in equation 1

163=\sigma_o+\dfrac{394.53}{\sqrt 35}

\sigma_o\ =96.31\ MPa

\sigma_y=96.31+\dfrac{394.53}{\sqrt d}

Now when d= 12 μm

\sigma_y=96.31+\dfrac{394.53}{\sqrt 12}

\sigma_y\ =210.2\ MPa

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