Chemistry, 29.01.2020 01:52 hartmat

Ablock weighing 8.5 n requires a force of 2 nto push it along at constant velocity. what is the coefficient of friction for thesurface?

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Answer from: Vampfox

The coefficient of friction of the surface μ = 0.235

Explanation:

Values of the given variables:

Weight of block = 8.5 N

Pushing force F = 2 N

velocity = constant

Solution:

Newton's Second Law of motion states that acceleration of an object depends on the mass of the object and the force applied

F = m a

Since v =constant, a = 0. Thus

Net force Fnet = 0.

Now F_net is the sum of the pushing force and the frictional force along the horizontal, which opposes the motion:

Fnet = Fpushing - Ffrictional_force

Given that Fnet = 0, Fpushing = Ffrictional_force. and Ffrictional_force = 2 N

Equation of friction is given by

Ffrictional_force = μ*W

2 = μ*8.5

μ = 2/8.5 = 0.235

Answer from: Quest

No table found. the question asks for covalent bond based on electronegativies

Answer from: Quest

Answerthis solution will not form a precipitateexplanationsolubility equilbrium: pbf₂(s) ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is $k_{sp} = [\text{pb}^{2+}][\text{f}^- ]^2 = 4 \times 10^{-8}$ (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as $q = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}}$ then, we can interpret q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product: $\begin{aligned} q & = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}} \\ & = (1.9 \times 10^{-3} ) (3.8\times 10^{3})^2 \\ & = 2.7 \times 10^{-8} \end{aligned}$ since q = 2.7 × 10⁻⁸ < ksp = 4 × 10⁻⁸, a precipitate will not form.

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Ablock weighing 8.5 n requires a force of 2 nto push it along at constant velocity. what is the coef...

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