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Ksju [112]
1 year ago
6

You probably noticed that the simulation returned varying values for the magnetic field. What other source(s) of error could occ

ur from the simulation
Physics
1 answer:
romanna [79]1 year ago
5 0

Physical and chemical laboratory experiments include three primary sources of error: systematic error, random error and human error. this simulation shows systematic error.

Systematic error applies to repeated measuring of the same object under repeated conditions of measurement. The amount of the error value is either positive or negative; which is called the fixed system error in laboratory experiments and laboratory tests.

Human error in laboratory experiments and lab tests primarily refers to the mistake in physical and chemical inspection phase caused by the factors of the inspector.

Random error is error caused by unknown factors. This error poses erratic changes at random, primarily due to a variety of small, independent, and accidental factors. The random error is atypical from the surface.

Learn more about systematic error here-

brainly.com/question/14149934

#SPJ4

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Power is calculated as work per unit time, and work in turn is calculated as force multiplied by distance. In this case, the force required is equivalent to the weight of the barbell multiplied by acceleration due to gravity.
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Measure the change in volume of the solution.

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Scientists want to place a telescope on the moon to improve their view of distant planets. The telescope weighs 200 pounds on Ea
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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
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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.
  • 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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What is a circuit. .
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