Answer:
2.16×10⁻⁶ N
Explanation:
Applying,
F = kqq'/r² (coulomb's Law)....................... Equation 1
Where F = electrostatic force, k = coulomb's constant, q = charge on the styrofoam, q' = charge on the grain of salt, r = distance between the charges.
From the question,
Given: q = 0.002 mC = 2.0×10⁻⁶ C, q' = 0.03 nC = 3.0×10⁻¹¹ C, r = 0.5 m
Constant: k = 8.99×10⁹ Nm²/C²
Substitute these values into equation 1
F = (2.0×10⁻⁶)(3.0×10⁻¹¹)(8.99×10⁹)/0.5²
F = 2.16×10⁻⁶ N
Prevailing definitions of climate are not much different from “the climate is what you expect, the weather is what you get”. Using a variety of sources including reanalyses and paleo data, and aided by notions and analysis techniques from Nonlinear Geophysics, we argue that this dictum is fundamentally wrong. <span>In addition to the weather and climate, there is a qualitatively distinct intermediate regime extending over a factor of ≈ 1000 in scale.Climate changes is projected to affect individual organisms, populations, ... Overall, there is a strong correlation between topographic slope and velocity from ... the ecosystems they live in—will adapt to these changes, or if they even can.</span>
True, because water balance is the balance between intake and output
But the fact is that an accelerating object is an object that is changing it’s velocity.. for this reason , it can be safely concluded that an object moving in a circle at constant speed is indeed accelerating. It is accelerating because the direction of the velocity vector is changing .
The amount of work done in emptying the tank by pumping the water over the top edge is 163.01* 10³ ft-lbs.
Given that, the tank is 8 feet across the top and 6 feet high
By the property of similar triangles, 4/6 = r/y
6r = 4y
r = 4/6*y = 2/3*y
Each disc is a circle with area, A = π(2/3*y)² = 4π/9*y²
The weight of each disc is m = ρw* A
m = 62.4* 4π/9*y² = 87.08*y²
The distance pumped is 6-y.
The work done in pumping the tank by pumping the water over the top edge is
W = 87.08 ∫(6-y)y² dy
W = 87.08 ∫(6y³ - y²) dy
W = 87.08 [6y⁴/4 - y³/3]
W = 87.08 [3y⁴/2- y³/3]
The limits are from 0 to 6.
W = 87.08 [3*6⁴/2 - 6³/3] = 87.08* [9*6³ - 2*36] = 87.08(1872) = 163013.76 ft-lbs
The amount of work done in emptying the tank by pumping the water over the top edge is 163013.76 ft-lbs.
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