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2 years ago

17. A 25 kg block is initially at rest on a rough, horizontal surface. A horizontal force of 75 N is required to set

Physics

1 answer:

lapo4ka [179]2 years ago

4 0

Answer:

0.30581

0.24464

Explanation:

$\mu_s$ = Coefficient of static friction

$\mu_k$ = Coefficient of kinetic friction

$F_f$ = 75 N

$F_k$ = 60 N

Normal force

$F_n=mg\\\Rightarrow F_n=25\times 9.81\\\Rightarrow F_n=245.25\ N$

Frictional force

$F_f=\mu_sF_n\\\Rightarrow \mu_s=\frac{F_f}{F_n}\\\Rightarrow \mu_s=\frac{75}{245.25}\\\Rightarrow \mu_s=0.30581$

The coefficient of static friction is 0.30581

Kinetic force

$F_k=\mu_kF_n\\\Rightarrow \mu_k=\frac{F_k}{F_n}\\\Rightarrow \mu_s=\frac{60}{245.25}\\\Rightarrow \mu_s=0.24464$

The coefficient of kinetic friction is 0.24464

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slega [8]

Answer:

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Its cell potential would be equal to $0.339 - (-0.130) = \rm 0.469\; V$ .

The maximum amount of electrical energy possible (under standard conditions) is equal to the free energy of this reaction:

$\Delta G^{\circ} = n \cdot F \cdot E^{\circ} (\text{cell})$ ,

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$n$ is the number moles of electrons transferred for each mole of the reaction. In this case the value of $n$ is $2$ as in the half-reactions.
$F$ is Faraday's Constant (approximately $96485.33212\; \rm C \cdot mol^{-1}$ .)

$\begin{aligned}\Delta G^{\circ} &= n \cdot F \cdot E^{\circ} (\text{cell})\cr &= 2\times 96485.33212 \times (0.339 - (-0.130)) \cr &\approx 9.0 \times 10^{4} \; \rm J \cr &= 90\; \rm kJ\end{aligned}$ .

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