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vfiekz [6]
3 years ago
12

An electron (q=-1.602×10-19C) is placed .03m away from spherical object with a net charge of -7.2 C.

Physics
1 answer:
vovangra [49]3 years ago
8 0

Answer:

Explanation:

electric field at the location of electron

= 9 x 10⁹ x 7.2 / .03²

= 72 x 10¹² N/C

force on electron = electric field x charge on electron

= 72 x 10¹² x 1.6 x 10⁻¹⁹

= 115.2 x 10⁻⁷ N .

C )

work done = charge on electron x potential difference at two points

potential at .03 m

= 9 x 10⁹ x 7.2 / .03

= 2.16 x 10¹² V

potential at .001 m

= 9 x 10⁹ x 7.2 / .001

= 64.8 x 10¹² V

potential difference = (64.8 - 2.16 )x 10¹² V

= 62.64 x 10¹² V  .

work done = 62.64 x 10¹² x 1.6 x 10⁻¹⁹

= 100.224 x 10⁻⁷ J .

D )

There will be no change in the magnitude of force on positron except that the direction of force will be reversed . In case of electron , there will be repulsion and in case of positron , there will be attraction .

Work done in case of electron will be positive and work done in case of positron will be negative .

electric field due to charge will be same in both the cases .

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Answer: 0.2  hours

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Fron this we can obtain:

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4 0
3 years ago
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iVinArrow [24]
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5 0
3 years ago
Read 2 more answers
An object weighing 1.840 kg has a volume of 0.0015 m3. What is the density of the object in g/cm3?
olga_2 [115]

Answer:

1.22gcm³

Explanation:

D = mass/ volume

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1m³= 1000,000cm³.

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X(cm³) = 1000,000×0.0015

X(cm³)= 1500.

Since density is mass/volume, now impute your data's

D=m/v

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Answer:

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Explanation:

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3 years ago
How to know which side is positive or negative on resistors.
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Answer:

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i hope it helped :)

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