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dlinn [17]
3 years ago
3

The electric force between a negatively charged paint droplet and a positively charged automobile body is increased by a factor

of two, but the charges on each remain constant. how has the distance between the two changed? (assume that the charge on the automobile is located at a single point.)
Physics
1 answer:
Gnom [1K]3 years ago
3 0
The electric force between the two objects is given by
F= k_e  \frac{q_1 q_2}{r^2}
where 
k_e is the Coulomb's constant
q_1 and q_2 are the charges of the two objects
r is the distance between the two objects

in the problem, the charges q1 and q2 remain constant, so the only quantity that changes is r, the distance. We can see that the force is proportional to \frac{1}{r^2}: this means that the distance r is proportional to
r \sim  \frac{1}{\sqrt{F}}
So, if the force has increased by a factor 2: F' = 2F, the new distance is
r'= \frac{1}{\sqrt{2F}} = \frac{r}{\sqrt{2} }

so, the distance has decreased by a factor \sqrt{2}.
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Answer: The correct statement is:

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Explanation:

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RESISTORS are said to be in parallel when two or more resistance or conductors are connected to common terminals so that the potential difference ( voltage) across each conductor IS THE SAME but with different current flow through each of them. Also, Individual resistances diminish to equal a smaller total resistance rather than add to make the total.

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8.67807 N

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Explanation:

m = Mass of shark = 92 kg

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F_n=mg-V_s\rho_{se}g\\\Rightarrow F_n=mg-\frac{m}{\rho_{sh}}\rho_{se}g\\\Rightarrow F_n=92\times 9.81-\frac{92}{1040}\times 1030\times 9.81\\\Rightarrow F_n=8.67807\ N

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F_n=mg-V_s\rho_{f}g\\\Rightarrow F_n=mg-\frac{m}{\rho_{sh}}\rho_{f}g\\\Rightarrow F_n=92\times 9.81-\frac{92}{1040}\times 1000\times 9.81\\\Rightarrow F_n=34.7123\ N

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