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Nostrana [21]
3 years ago
11

Which of the following materials is a good conductor of electric current??? Anyone the options are A.) a rubber tire B.) a silve

r spoon C.) a lightbulb filament D.) a glass pane
Physics
1 answer:
bekas [8.4K]3 years ago
6 0
A silver spoon (B.) A silver spoon is metal so it would be the best answer out of the four. a rubber tire is a good insulator but not a good conductor. Thank you!
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Franklin needs to ship a box with FedEx. In order to calculate his shipping costs, he needs to measure the mass of the package.
balandron [24]

Answer:

meter

Explanation:

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What is the avarage velocity of a car that travels 30 kilometers due west in 0.5 hours
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Which quantity measures the maximum distance an object on a spring moves from the equilibrium position during oscillation?
nasty-shy [4]
The appropriate response is amplitude. A measure of its change over a solitary period. There are different meanings of plentifulness, which are all elements of the extent of the distinction between the variable's outrageous qualities. In more seasoned writings, the stage is now and again called the adequacy.
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xConsider the following reduction potentials: Cu2+ + 2e– Cu E° = 0.339 V Pb2+ + 2e– Pb E° = –0.130 V For a galvanic cell employi
slega [8]

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.
  • 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}.

5 0
3 years ago
When a condenser discharges electricity, the instantaneous rate of change of the voltage is proportional to the voltage in the c
e-lub [12.9K]

Answer:

460.52 s

Explanation:

Since the instantaneous rate of change of the voltage is proportional to the voltage in the condenser, we have that

dV/dt ∝ V

dV/dt = kV

separating the variables, we have

dV/V = kdt

integrating both sides, we have

∫dV/V = ∫kdt

㏑(V/V₀) = kt

V/V₀ = e^{kt}

Since the instantaneous rate of change of the voltage is -0.01 of the voltage dV/dt = -0.01V

Since dV/dt = kV

-0.01V = kV

k = -0.01

So, V/V₀ = e^{-0.01t}

V = V₀e^{-0.01t}

Given that the voltage decreases by 90 %, we have that the remaining voltage (100 % - 90%)V₀ = 10%V₀ = 0.1V₀

So, V = 0.1V₀

Thus

V = V₀e^{-0.01t}

0.1V₀ = V₀e^{-0.01t}

0.1V₀/V₀ = e^{-0.01t}

0.1 = e^{-0.01t}

to find the time, t it takes the voltage to decrease by 90%, we taking natural logarithm of both sides, we have

㏑(0.01) = -0.01t

So, t = ㏑(0.01)/-0.01

t = -4.6052/-0.01

t = 460.52 s

3 0
3 years ago
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