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gladu [14]
3 years ago
5

In a voltaic cell, electrons flow from the (positive/negative) to the (positive/negative) terminal.

Physics
2 answers:
oksano4ka [1.4K]3 years ago
6 0

Explanation :

Voltaic cells are an electro chemical cell in which chemical reactions occurs to produce electrical energy. Its parts are :

(i) At anode oxidation occurs.

(ii) At cathode reduction occurs.

(iii) A salt bridge.

(iv) External circuit.

(v) Load.

Suppose a voltaic cell is made up of ZnSO_4 and CuSO_4. Here zinc can lose electron more rapidly as that of copper. Oxidation of zinc occurs and is called as anode. So, this is a negative terminal.

On the other hand, copper receives electron and hence become cathode. So, this is a positive terminal.

Hence, electrons flow form Zinc to copper i.e. from negative terminal to positive terminal.

Usimov [2.4K]3 years ago
6 0

Answer:

^ Shirleywashington is correct. The rating of her answer is misleading, but the answer is negative to positive.

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Two identical small metal spheres with q1 &gt; 0 and |q1| &gt; |q2| attract each other with a force of magnitude 72.1 mN when se
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1) +2.19\mu C

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F=k\frac{q_1 q_2}{r^2} (1)

where

k is the Coulomb's constant

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r is the separation between the charges

When the two spheres are brought in contact with each other, the charge equally redistribute among the two spheres, such that each sphere will have a charge of

\frac{Q}{2}

where Q is the total charge between the two spheres.

So we can actually rewrite the force as

F=k\frac{(\frac{Q}{2})^2}{r^2}

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We can solve the equation for Q:

Q=2\sqrt{\frac{Fr^2}{k}}=2\sqrt{\frac{(0.02163)(1.41)^2}{8.98755\cdot 10^9}}}=4.37\cdot 10^{-6} C

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F=k \frac{q_1 (Q-q_1)}{r^2}

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Fr^2=k (q_1 Q-q_1^2})\\kq_1^2 -kQ q_1 +Fr^2 = 0

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q_1 = 6.70\cdot 10^{-6} C\\q_2 = -2.34\cdot 10^{-6} C

Which correspond to the values of the two charges. Therefore, the initial charge q1 on the first sphere is

q_1 = +6.70 \mu C

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