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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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When driving at night, only use your high-beam headlights___
zzz [600]

Answer:

B

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An astronaut named Speedo takes a trip to Sirius, which is located a distance of 8 lightyears from the Earth. If the spaceship m
Softa [21]

Answer:

<em>a. 10 years</em>

<em>b. 6 years</em>

<em>c. 12 years behind</em>

<em></em>

Explanation:

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distance of sirius from earth = 8 lightyears = 8 x c x 60 x 60 x 24 x 365 = 252288000c meters

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time taken to reach sirius = distance covered divided by speed

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<em>1 year = 60 x 60 x 24 x 365 sec = 31536000 sec</em>

315360000 sec = 315360000/31536000 = <em>10 yrs</em>

<em>the time as observed by Goslo will be shorter than the real time as measured within the ship due to relativistic effects. using the relativistic relation for time dilation we have</em>

t' = t\sqrt{1 - \frac{v^{2} }{c^{2} } }

since v = 0.8c

v/c = 0.8

where t' is the relativistic time as observed by Goslo,

t is the time within the ship

t' = 10 x \sqrt{1 - 0.8^{2}  }  = <em>6 yrs</em>

<em></em>

If speedo turns around immediately after reaching  sirius, his total distance will be 252288000c x 2 = 50457600c meter

time taken = 50457600c/0.8 = 630720000 sec = 20 yrs

time within the ship will be calculated from time dilation

t' = t\sqrt{1 - \frac{v^{2} }{c^{2} } }

t' = 20 x  \sqrt{1 - 0.8^{2}  }  = 378432000 sec = <em>12 years behind</em>

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jeka57 [31]

Answer:

A

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6 0
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