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elixir [45]
3 years ago
10

Good morning! someone please answer this, ill give you brainliest and your earning 50 points.

Physics
1 answer:
wariber [46]3 years ago
8 0

One step of copper mining requires blasting ore to be able to blasted and extracted from mine.

  • <em>This</em><em> </em><em>process</em><em> </em><em>causes</em><em> </em><em>harm</em><em> </em><em>to</em><em> </em><em>the</em><em> </em><em>Geosphere</em><em> </em><em>(</em><em>Lithosphere</em><em>)</em>
  • <u>In</u><u> </u><u>this</u><u> </u><u>step</u><u>,</u><u> </u><u>they</u><u> </u><u>blast</u><u> </u><u>the</u><u> </u><u>mines</u><u> </u><u>inside</u><u> </u><u>the</u><u> </u><u>earth</u><u>.</u><u> </u><u>Which</u><u> </u><u>causes</u><u> </u><u>destruction</u><u> </u><u>and</u><u> </u><u>harm</u><u> </u><u>to</u><u> </u><u>the</u><u> </u><u>geosphere</u><u>.</u><u> </u>
  • <u>We</u><u> </u><u>know</u><u> </u><u>that</u><u>,</u><u> </u><u>land</u><u> </u><u>comes</u><u> </u><u>under</u><u> </u><u>G</u><u>e</u><u>osphere</u><u> </u><u>also</u><u> </u><u>known</u><u> </u><u>as</u><u> </u><u>Lithosphere</u><u>.</u>

Mercury is used to recover minute pieces of gold that is mixed in soil and sediment. Sometimes Mercury can be lost in this process resulting in widespread contamination of local rivers and lakes.

  • <em>This</em><em> </em><em>causes</em><em> </em><em>harm</em><em> </em><em>to</em><em> </em><em>the</em><em> </em><em>Hydrosphere</em><em>.</em>
  • <u>The</u><u> </u><u>portion</u><u> </u><u>of</u><u> </u><u>earth</u><u> </u><u>covered</u><u> </u><u>with</u><u> </u><u>water</u><u> </u><u>is</u><u> </u><u>known</u><u> </u><u>as</u><u> </u><u>hydrosphere</u><u>.</u>
  • <u>Since</u><u>,</u><u> </u><u>The</u><u> </u><u>Mercury</u><u> </u><u>is</u><u> </u><u>getting</u><u> </u><u>mixed</u><u> </u><u>with</u><u> </u><u>the</u><u> </u><u>water</u><u>.</u><u> </u><u>It</u><u> </u><u>is</u><u> </u><u>polluting</u><u> </u><u>the</u><u> </u><u>water</u><u> </u><u>that</u><u> </u><u>is</u><u> </u><u>the</u><u> </u><u>hydrosphere</u><u>.</u>

Hundreds of tons of rock are unearthed, moved and crushed in mining operation significantly increasing the amount of dust and particulates in the air.

  • <em>This</em><em> </em><em>causes</em><em> </em><em>harm</em><em> </em><em>to</em><em> </em><em>the</em><em> </em><em>Atmosphere</em><em>.</em>
  • <u>The</u><u> </u><u>particles</u><u> </u><u>of</u><u> </u><u>dust</u><u> </u><u>get</u><u> </u><u>increased</u><u> </u><u>in</u><u> </u><u>air</u><u> </u><u>and</u><u> </u><u>eventually</u><u> </u><u>gets</u><u> </u><u>mixed</u><u> </u><u>and</u><u> </u><u>spread</u><u> </u><u>everywhere</u><u>.</u>
  • <u>The</u><u> </u><u>air</u><u> </u><u>is</u><u> </u><u>getting</u><u> </u><u>polluted</u><u> </u><u>which</u><u> </u><u>means</u><u> </u><u>the</u><u> </u><u>atmosphere</u><u> </u><u>is</u><u> </u><u>getting</u><u> </u><u>harmed</u><u>.</u>

Strip mining (also known as mountaintop removal) involves scraping away earth and rocks to get to coal buried near the surface. This often results in the destruction of ecosystems.

  • <em>This</em><em> </em><em>causes</em><em> </em><em>harm</em><em> </em><em>to</em><em> </em><em>the</em><em> </em><em>Biosphere</em><em>.</em>
  • <u>The</u><u> </u><u>biosphere</u><u> </u><u>consists</u><u> </u><u>of</u><u> </u><u>plants</u><u> </u><u>and</u><u> </u><u>animals</u><u>.</u><u> </u><u>And</u><u> </u><u>these</u><u> </u><u>plants</u><u> </u><u>and</u><u> </u><u>animals</u><u> </u><u>collectively</u><u> </u><u>form</u><u> </u><u>a</u><u> </u><u>habitat</u><u> </u><u>called</u><u> </u><u>ecosystem</u><u>.</u>
  • <u>When</u><u> </u><u>a</u><u> </u><u>ecosystem</u><u> </u><u>is</u><u> </u><u>destroyed</u><u>,</u><u> </u><u>the</u><u> </u><u>biosphere</u><u> </u><u>is</u><u> </u><u>harmed</u><u>.</u>
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PLZ HELP ON #22-26!!!! <br><br>Please explain why and how you got your answer.
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22. a - (vf^2 - vi^2)/(2d) 
a = (0 - 23^2)/(170) 
a = -3.1 m/s^2

23. Find the time (t) to reach 33 m/s at 3 m/s^2
33-0/t = 3 
33 = 3t 
t = 11 sec to reach 33 m/s^2
Find the av velocuty: 33+0/2 = 16.5 m/s
Dist = 16.5 * 11 = 181.5 meters to each 33m/s speed. Runway has to be at least this long. 

24. The sprinter starts from rest. The average acceleration is found from: 
(Vf)^2 = (Vi)^2 = 2as ---> a = (Vf)^2 - (Vi)^2/2s = (11.5m/s)^2-0/2(15.0m) = 4.408m/s^2 estimated: 4.41m/s^2
The elapsed time is found by solving
Vf = Vi + at ----> t = vf-vi/a = 11.5m/s-0/4.408m/s^2 = 2.61s

25. Acceleration of car = v-u/t = 0ms^-1-21.0ms^-1/6.00s = -3.50ms^-2
S = v^2 - u^2/2a = (0ms^-1)^2-(21.0ms^-1)^2/2*-3.50ms^-2 = 63.0m 

26. Assuming a constant deceleration of 7.00 m/s^2
final velocity, v = 0m/s 
acceleration, a = -7.00m/s^2
displacement, s - 92m 
Using v^2 = u^2 - 2as 
0^2 - u^2 + 2 (-7.00) (92) 
initial velocity, u = sqrt (1288) = 35.9 m/s 
This is the speed pf the car just bore braking. 

I hope this helps!! 

5 0
3 years ago
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