1. Which of the following materials would likely create the strongest, most stable slope?
2 answers:
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Answer:
see explanation
Explanation:
You are missing the chart with the rates and time to do this, however, I wll do it with a similar exercise here, and you only need to replace the procedure with your data:
See the attached table.
From the left we have:
r = 1/2 (50 + 48 + 46 + 44 + 42 + 40) = 135 L/min
From the right we have:
r = 1/2 (48 + 46 +44 + 42 + 40 + 38) = 129 L/min.
And this should be the correct answer. Watch your chart and replace if it's neccesary.
Answer:
electricfield at (0,0,0) is Et = 2 k q / a²
Explanation:
For the first part see the diagram , the field lines start from the positive charge and reach the negative charge, notice that no line should cross, some lines go to infinity
For the second part we use that the electric field is a vector quantity and therefore we add the field of each charge, using the equation
E = k q / r²
Point (0,0,0)
We calculate for the charge -q which is at a distance R = a
E1 = k (-q) / a²
E1 = - kq / a²
As the test charge is positive in the field it goes to the left, attractive force
We calculate for the charge that is also at R = a
E2 = k q / a²
This field goes to the left, repulsive force
We find the total electric field
Et = E1 + E2
Et = kq / a² + kq / a²
Et = 2 k q / a²
Point (0,0, R)
We use the same equations, but with another distance, for the charge -q the distance is R = R+a and for the charge + q the distance is R = R-a
E1 = k q / (R + a)²
E2 = kq / (R-a)²
Et = kq [1 / (R + a)² + 1 / (R-a)²]
Et= kq {[(R-a)² + (R + a)²] / [(R + a)² (R-a)²]}
Et= kq {2 (R² + a²) / [(R + a)² (R-a)²]}
If we use the condition that R> a we can despise in the patents "a"
(R² + a²) = R² (1+ a² / R²) ≈ R²
(R + a)² = R² (1 + a / R)² ≈ R²
(R- a)² = R² (1-a / R)² ≈ R²
Substituting in the total electric field
Et = kq {2 R²) / [R²R²]}
Et =kq 2 / R²
