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lidiya [134]
4 years ago
5

In a car moving at constant acceleration, you travel 290 m between the instants at which the speedometer reads 40 km/h and 80 km

/h . Part a part complete how many seconds does it take you to travel the 290 m ?
Physics
1 answer:
liubo4ka [24]4 years ago
4 0

<u>Answer:</u>

 Car will take 17.36 seconds to travel the 290 meter.

<u>Explanation:</u>

   We know the equation of motion v^2 =u^2+2as , where s is the displacement, u is the initial velocity, a is the acceleration and v is the final velocity.

  In this case initial velocity = 40 km/h = 11.11 m/s

                      Final velocity = 80 km/h = 22.22 m/s

                      Displacement = 290 m

   Substituting         

          22.22^2=11.11^2+2*a*290\\ \\ a = 0.64 m/s^2

 Now we have v = u+at, where t is the time taken.

   Substituting

            22.22 =11.11 + 0.64*t

                     t = 17.36 seconds

So car will take 17.36 seconds to travel the 290 meter.

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3 years ago
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Answer:

Approximately \rm 90\; kJ.

Explanation:

Cathode is where reduction takes place and anode is where oxidation takes place. The potential of a electrochemical reaction (E^{\circ}(\text{cell})) is equal to

E^{\circ}(\text{cell}) = E^{\circ}(\text{cathode}) - E^{\circ}(\text{anode}).

There are two half-reactions in this question. \rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu and \rm Pb^{2+} + 2\,e^{-} \rightleftharpoons Pb. Either could be the cathode (while the other acts as the anode.) However, for the reaction to be spontaneous, the value of E^{\circ}(\text{cell}) should be positive.

In this case, E^{\circ}(\text{cell}) is positive only if \rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu is the reaction takes place at the cathode. The net reaction would be

\rm Cu^{2+} + Pb \to Cu + Pb^{2+}.

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}),

where

  • 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.
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\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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