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aleksley [76]
2 years ago
9

A 2.03 kg book is placed on a flat desk. Suppose the coefficient of static friction between the book and the desk is 0.602 and t

he coefficient of kinetic friction is 0.327 . How much force is needed to begin moving the book
Physics
1 answer:
diamong [38]2 years ago
5 0

Answer:

11.98 N

Explanation:

Normal force =   mg =  2.03 * 9.81

coeff of static friction must be overcome for the book to begin moving

       .602 = F / (2.03 * 9.81)   = 11.98  N

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Zero or +2

Explanation:

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What happens to light waves from a star as the star moves away from Earth?
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<h2>Answer: Light waves have a redshift due to the Doppler effect </h2>

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3 years ago
ome metal oxides can be decomposed to the metal and oxygen under reasonable conditions. 2 Ag2O(s) → 4 Ag(s) + O2(g) Thermodynami
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Answer : The values of \Delta H^o,\Delta S^o\text{ and }\Delta G^o are 62.2kJ,132.67J/K\text{ and }22.66kJ respectively.

Explanation :

The given balanced chemical reaction is,

2Ag_2O(s)\rightarrow 4Ag(s)+O_2(g)

First we have to calculate the enthalpy of reaction (\Delta H^o).

\Delta H^o=H_f_{product}-H_f_{product}

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where,

\Delta H^o = enthalpy of reaction = ?

n = number of moles

\Delta H_f^0 = standard enthalpy of formation

Now put all the given values in this expression, we get:

\Delta H^o=[4mole\times (0kJ/mol)+1mole\times (0kJ/mol)}]-[2mole\times (-31.1kJ/mol)]

\Delta H^o=62.2kJ=62200J

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction (\Delta S^o).

\Delta S^o=S_f_{product}-S_f_{product}

\Delta S^o=[n_{Ag}\times \Delta S_f^0_{(Ag)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]-[n_{Ag_2O}\times \Delta S_f^0_{(Ag_2O)}]

where,

\Delta S^o = entropy of reaction = ?

n = number of moles

\Delta S_f^0 = standard entropy of formation

Now put all the given values in this expression, we get:

\Delta S^o=[4mole\times (42.55J/K.mole)+1mole\times (205.07J/K.mole)}]-[2mole\times (121.3J/K.mole)]

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Now we have to calculate the Gibbs free energy of reaction (\Delta G^o).

As we know that,

\Delta G^o=\Delta H^o-T\Delta S^o

At room temperature, the temperature is 298 K.

\Delta G^o=(62200J)-(298K\times 132.67J/K)

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Therefore, the values of \Delta H^o,\Delta S^o\text{ and }\Delta G^o are 62.2kJ,132.67J/K\text{ and }22.66kJ respectively.

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