Acid mine drainage is the formation and movement of highly acidic water rich in heavy metals. This acidic water forms through the chemical reaction of surface water (rainwater, snowmelt, pond water) and shallow subsurface water with rocks that contain sulfur-bearing minerals, resulting in sulfuric acid.
The answer is D-all choices
Answer:
critical stress = 595 MPa
Explanation:
given data
fracture toughness = 74.6 MPa-![\sqrt{m}](https://tex.z-dn.net/?f=%5Csqrt%7Bm%7D)
crack length = 10 mm
f = 1
solution
we know crack length = 10 mm
and crack length = 2a as given in figure attach
so 2a = 10
a = 5 mm
and now we get here with the help of plane strain condition , critical stress is express as
critical stress =
......................1
put here value and we get
critical stress =
critical stress = 595 MPa
so here stress is change by plane strain condition because when plate become thinner than condition change by plane strain to plain stress.
plain stress condition occur in thin body where stress through thickness not vary by the thinner section.
Answer:
Explanation:
Using the kinematics equation
to determine the velocity of car B.
where;
initial velocity
= constant deceleration
Assuming the constant deceleration is = -12 ft/s^2
Also, the kinematic equation that relates to the distance with the time is:
![S = d + v_ot + \dfrac{1}{2}at^2](https://tex.z-dn.net/?f=S%20%3D%20d%20%2B%20v_ot%20%2B%20%5Cdfrac%7B1%7D%7B2%7Dat%5E2)
Then:
![v_B = 60-12t](https://tex.z-dn.net/?f=v_B%20%3D%2060-12t)
The distance traveled by car B in the given time (t) is expressed as:
![S_B = d + 60 t - \dfrac{1}{2}(12t^2)](https://tex.z-dn.net/?f=S_B%20%3D%20d%20%2B%2060%20t%20-%20%5Cdfrac%7B1%7D%7B2%7D%2812t%5E2%29)
For car A, the needed time (t) to come to rest is:
![v_A = 60 - 18(t-0.75)](https://tex.z-dn.net/?f=v_A%20%3D%2060%20-%2018%28t-0.75%29)
Also, the distance traveled by car A in the given time (t) is expressed as:
![S_A = 60 * 0.75 +60(t-0.75) -\dfrac{1}{2}*18*(t-0.750)^2](https://tex.z-dn.net/?f=S_A%20%3D%2060%20%20%2A%200.75%20%2B60%28t-0.75%29%20-%5Cdfrac%7B1%7D%7B2%7D%2A18%2A%28t-0.750%29%5E2)
Relating both velocities:
![v_B = v_A](https://tex.z-dn.net/?f=v_B%20%3D%20v_A)
![60-12t = 60 - 18(t-0.75)](https://tex.z-dn.net/?f=60-12t%20%3D%2060%20-%2018%28t-0.75%29)
![60-12t =73.5 - 18t](https://tex.z-dn.net/?f=60-12t%20%3D73.5%20-%2018t)
![60- 73.5 = - 18t+ 12t](https://tex.z-dn.net/?f=60-%2073.5%20%3D%20-%2018t%2B%2012t)
![-13.5 =-6t](https://tex.z-dn.net/?f=-13.5%20%3D-6t)
t = 2.25 s
At t = 2.25s, the required minimum distance can be estimated by equating both distances traveled by both cars
i.e.
![S_B = S_A](https://tex.z-dn.net/?f=S_B%20%3D%20S_A)
![d + 60 t - \dfrac{1}{2}(12t^2) = 60 * 0.75 +60(t-0.75) -\dfrac{1}{2}*18*(t-0.750)^2](https://tex.z-dn.net/?f=d%20%2B%2060%20t%20-%20%5Cdfrac%7B1%7D%7B2%7D%2812t%5E2%29%20%3D%2060%20%20%2A%200.75%20%2B60%28t-0.75%29%20-%5Cdfrac%7B1%7D%7B2%7D%2A18%2A%28t-0.750%29%5E2)
![d + 60 (2.25) - \dfrac{1}{2}(12*(2.25)^2) = 60 * 0.75 +60((2.25)-0.75) -\dfrac{1}{2}*18*((2.25)-0.750)^2](https://tex.z-dn.net/?f=d%20%2B%2060%20%282.25%29%20-%20%5Cdfrac%7B1%7D%7B2%7D%2812%2A%282.25%29%5E2%29%20%3D%2060%20%20%2A%200.75%20%2B60%28%282.25%29-0.75%29%20-%5Cdfrac%7B1%7D%7B2%7D%2A18%2A%28%282.25%29-0.750%29%5E2)
d + 104.625 = 114.75
d = 114.75 - 104.625
d = 10.125 ft
Answer:
The Space Needle is a cut away with minimal residual deflection due to load transfer.