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kvasek [131]
3 years ago
5

Two uncharged metal spheres, #1 and #2, are mounted on insulating support rods. A third metal sphere, carrying a positive charge

, is then placed near #2. Now a copper wire is momentarily connected between #1 and #2 and then removed. Finally, sphere #3 is removed. In this final state
Physics
1 answer:
tatuchka [14]3 years ago
5 0

Answer:

The question is incomplete as some details are missing, here is the details ;

In this final state

a) spheres #1 and #2 both carry negative charge.

b) sphere #1 carries negative charge and #2 carries positive charge.

c) spheres #1 and #2 are still uncharged.

d) sphere #1 carries positive charge and #2 carries negative charge.

e) spheres #1 and #2 both carry positive charge.

Explanation:

From the concept of electrostatics, if a positively charged sphere is brought close to #2, there will be attraction of the opposite charges(-ve) towards it.

Now connecting a copper wire between #1 and #2, opposite charges will flow from #1 towards #2. disconnecting the copper wire makes #1 to be positively charged and #2 to be negatively charge and from the laws of attraction ; Like charges repel and unlike charges attract. the correct option is d.

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4.2 km

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Using sine rule:

\frac{b}{sin(B)}=\frac{c}{sin(C)}  \\\\\frac{8}{sin(120)}=\frac{5}{sin(C)}  \\\\sin(C)=\frac{5*sin(120)}{8} =0.5412\\\\C=sin^{-1}(0.5412)=32.8^o\\\\

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∠A + 120 + 32.8 = 180

∠A = 27.2°

\frac{b}{sin(B)}=\frac{a}{sin(A)}  \\\\\frac{8}{sin(120)}=\frac{a}{sin(27.2)}  \\\\a=\frac{8*sin(27.2)}{sin(120)} \\\\a=4.2\ km

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3 years ago
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