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victus00 [196]
3 years ago
10

The three pairs of metal same-size spheres have different charges on their surfaces, as indicated. Each pair is brought together

, allowed to touch, and then separated. Rank from greatest to least the total amount of charge on the pairs of spheres after separation.
A) +6 +2
B) +6 -2
C) +6 0
Physics
1 answer:
vodomira [7]3 years ago
8 0

Answer:

A≥C>B

Explanation:

The total charge on the spheres depends on the pair attached. considering pair (A), the pair of +6 and +2 will produce a net charge +8 on pair A. Pair (B) has one positive and negative sphere of + 6 charge and other having -2 charge when brought near the net charge becomes +4. Pair C has a sphere having +6 charge and other is neutral i.e 0 charge when brought near the net charge on the pair becomes +6. In terms of ranking, pair A is the highest while pair B is the lowest.

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SHOW WORK
Helga [31]

Answer:

Follows are the solution to the given question:

Explanation:

For point a:

T= 2\pi \sqrt{\frac{m}{k}}\\\\k = \frac{4 \pi^2 m}{T^2}\\\\= \frac{4 \times (3.14)^2 \times 3}{2^2}\\\\=29.578 \ \frac{N}{m}\\\\

For Point b:

E=\frac{1}{2} m a^2 w^2\\\\

   =\frac{1}{2} \times m \times a^2 \times \frac{4\pi^2}{T^2}\\\\=\frac{1}{2} \times 3 \times (0.15)^2 \times \frac{4\times 3.14^2}{2^2}\\\\=0.332 \ J

For Point C:

V_{max}= a w

        = (0.15) \times \frac{2\pi}{T}\\\\= (0.15) \times \frac{2\times 3.14}{2}\\\\=0.471 \frac{m}{s}

For point D:

X= a \sin (wt+ \phi)\\\\0.91=0.15 \sin(\frac{2\pi}{T} \times t+\phi)\\\\0.91=0.15 \sin(\frac{2\times 3.14}{2} \times 0.5+\phi)\\\\0.60 = \sin(3.14 \times 0.5+\phi)\\\\0.60 = \sin(1.57+\phi)\\\\1.57 +\phi =\sin^{-1} 60^{\circ}\\\\1.57 +\phi = 36.86^{\circ}\\\\=35.29^{\circ}\\\\So, X=15 \sin(3.14t+35.29^{\circ}) \ cm

5 0
3 years ago
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