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antiseptic1488 [7]
3 years ago
10

Determine the field strength, E, experienced by a test charge, q, if a charge of 7.0 × 10-5 coulombs is placed on q and a force

of 5.2 N is applied.
Physics
2 answers:
kkurt [141]3 years ago
7 0
Formula for feild strength= F/q
q=7.0^10-5 coulombs
F=5.2 N
E=5.2 / 7.0^10-5
E=
Evgesh-ka [11]3 years ago
5 0
E = F/q

   = 5.2 N / (7* 10⁻⁵) C

   = 7.429 * 10⁴ N/C
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Answer:

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Rotation at the equator = 10 hours 14 minutes

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Therefore;

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The differential rotation = 24 minutes = 24 minutes × 1 hour/(60 minutes) = 0.4 hours

The differential rotation = 0.4 hours

Therefore, the measured day at the higher altitude will be 0.4 longer than at the equator.

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masha68 [24]

Answer:

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ycow [4]
I believe that the answer to this would be B




Hope this helped
3 0
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) A striker can give the ball an initial speed of 30m/s. Within what two elevation angles must he kick
sveticcg [70]

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I find it convenient to use a graphing calculator to find solutions for problems of this sort. In the attachment, we have used x as the angle in degrees, and written the function so that x-intercepts are the solutions.

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