Physics, 07.09.2019 01:30 Kinglilray998

An electroscope has a charge of −2.0 µc while your finger has a charge of 2.7 µc. you touch the electroscope with your finger and discover that the charges of your finger and the electroscope are now the same. in this process, electrons were transferred from the electroscope to your finger. how many electrons were transferred in order for your finger and the electroscope to have the same charge?

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Answer from: nakarelinp0p303

1.39 × 10⁻¹³ electrons were transferred

Explanation:

Given:

Charge on electroscope = - 2.0 μC

charge on finger = 2.7 μC

Now,

The net charge when in contact

= Charge on electroscope + charge on finger

The net charge when in contact = - 2.0 μC + 2.7 μC = 0.7 μC

Also,

Charge on the electroscope and the finger is same

thus,

The charge on the electroscope = $\frac{\textup{Net charge}}{\textup{2}}$

The charge on the electroscope = $\frac{0.7}{\textup{2}}$

The charge on the electroscope = 0.35 μC

Now,

The change in charge on the electroscope

= Final charge - Initial charge on the electroscope

= 0.35 μC - (- 2.0 μC)

The change in charge on the electroscope = 2.35 μC

also,

Charge = Number of electrons × Charge on an electron

2.35 μC = Number of electron × 1.69 × 10⁻¹⁹ C

2.35 × 10⁻⁶ C = Number of electron × 1.69 × 10⁻¹⁹ C

Number of electron = $\frac{\textup{2.35\times10^{-6}}}{\textup{1.69\times10^{-19}}}$

Number of electron = 1.39 × 10⁻¹³ electrons were transferred.

Answer from: Quest

The answer is water molecules are split.cellular respiration and photosynthesis are complement reactions to each other. these are reverse reactions. in cellular respiration glucose and oxygen yield water and carbon dioxide. so, when respiration begins, glucose molecules are split. in the opposite reaction - in photosynthesis, carbon dioxide and water yield glucose and oxygen. thus, when respiration begins, water molecules are split.

Answer from: Quest

Answerthe energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.explanationwe define electric potential as the ratio electric potential energy to the charge, $v = \dfrac{e_p}{q}$ then the following gives us the potential difference, $\delta v = v_b - v_a$ $\begin{aligned} \delta v & = \dfrac{e_p_b}{q} - \dfrac{e_p_a}{q} \\ & = \dfrac{\delta e_p}{q} \end{aligned}$ the formula $\delta v = \delta e_p / q$ may be on your formula sheet.the change in electric potential energy would be the energy that would needed to move it from point a to point b. solving for that, we get $\delta e_p = q\delta v$ using q = 8.00 × 10⁻¹⁹ c and δv = 2.00 v, we get $\begin{aligned} \delta e_p & = (8.00\times10^{-19} \text{ c})(2.00\text{ v}) \\ & = 1.60\times 10^{-18}\text{ j} \end{aligned}$ the energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.

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An electroscope has a charge of −2.0 µc while your finger has a charge of 2.7 µc. you touch the elec...

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