Physics, 09.06.2021 19:50 Bassoonist

Which alkali metal is most reactive?

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relativism, roughly put, is the view that truth and falsity, right and wrong, standards of reasoning, and procedures of justification are products of differing conventions and frameworks of assessment and that their authority is confined to the context giving rise to them. more precisely, “relativism” covers views which maintain that—at a high level of abstraction—at least some class of things have the properties they have (e.g., beautiful, morally good, epistemically justified) not simpliciter, but only relative to a given framework of assessment (e.g., local cultural norms, individual standards), and correspondingly, that the truth of claims attributing these properties holds only once the relevant framework of assessment is specified or supplied. relativists characteristically insist, furthermore, that if something is only relatively so, then there can be no framework-independent vantage point from which the matter of whether the thing in question is so can be established.

relativism has been, in its various guises, both one of the most popular and most reviled philosophical doctrines of our time. defenders see it as a harbinger of tolerance and the only ethical and epistemic stance worthy of the open-minded and tolerant. detractors dismiss it for its alleged incoherence and uncritical intellectual permissiveness. debates about relativism permeate the whole spectrum of philosophical sub-disciplines. from ethics to epistemology, science to religion, political theory to ontology, theories of meaning and even logic, philosophy has felt the need to respond to this heady and seemingly subversive idea. discussions of relativism often also invoke considerations relevant to the very nature and methodology of philosophy and to the division between the so-called “analytic and continental” camps in philosophy. and yet, despite a long history of debate going back to plato and an increasingly large body of writing, it is still difficult to come to an agreed definition of what, at its core, relativism is, and what philosophical import it has. this entry attempts to provide a broad account of the many ways in which “relativism” has been defined, explained, defended and criticized.

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Answer from: Quest

pecialized keyboards

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the speed of the cart after 3 seconds of low fan speed is ⇒ 54.0 cm/s.

the speed of the cart after 5 seconds of medium fan speed is ⇒ 120.0 cm/s.

the speed of the cart after 2 seconds of high fan speed is ⇒ 64.0 cm/s.

just did the lab.

Answer from: Quest

Answer: a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same sign as the -4.0 µc charge.======electricstatic force between two point charges is given by $f = \dfrac{kq_1q_2}{r^2}$ where k is coulomb's constant, 8.99×10⁹ n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ = 6.0 µc, q₂ = -4.0 µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1 µc = 10⁻⁶ c) $f_{01} = \dfrac{kq_0q_1}{(\text{6.0 m})^2}$ the force from q₂ $f_{02} = \dfrac{kq_0q_2}{(\text{15 m})^2}$ the force from q₃ $f_{03} = \dfrac{kq_0q_1}{(\text{26 m})^2}$ the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃ $\begin{aligned} f_{net} & = 0\\ f_{01} + f_{02} + f_{03} & = 0 \\ \dfrac{kq_0q_1}{(\text{6.0 m})^2} + \dfrac{kq_0q_2}{(\text{15 m})^2} + \dfrac{kq_0q_3}{(\text{26 m})^2} & = 0 \\ kq_0 \left( \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2}\right) & = 0 \\ \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2} & = 0 \end{aligned}$ isolating: $\begin{aligned} \dfrac{q_3}{(\text{26 m})^2} & = -\dfrac{q_1}{(\text{6.0 m})^2} - \dfrac{q_2}{(\text{15 m})^2} \\ q_3 & = -(\text{26 m})^2\left(\dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2}\right) \\ q_3 & = -(\text{26 m})^2\left(\dfrac{6.0\times10^{-6} \text{ c}}{(\text{6.0 m})^2} + \dfrac{-4.0\times10^{-6} \text{ c}}{(\text{15 m})^2}\right) \\ q_3 & = -1.0064888 \times 10^{-4}\text{ c} \\ q_3 & = -1.0\times 10^{-4}\text{ c} \end{aligned}$ a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same as the -4.0 µc charge