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aniked [119]
3 years ago
9

At what distance from a 1.00*10^-5 C charge will the electric potential be 10000 V? (Unit = m)​

Physics
2 answers:
OlgaM077 [116]3 years ago
7 0

Answer:

9.0 m

Explanation:

Applying

V = kq/r......................... Equation 1

Where V = Electric potential, q = charge, r = distance, k = Coulomb's constant.

make r the subject of the equation

r = kq/V................... Equation 2

From the question,

Given: q = 1.00×10⁻⁵ C, V = 10000 V

Constant: k = 9.0×10⁹ N⋅m²/C²

Substitute these values into equation 2

r = (9.0×10⁹×1.00×10⁻⁵)/10000

r = 9.0 m

lozanna [386]3 years ago
6 0

Answer:

9.0

Explanation:

Correct for Acellus Students:)

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Explanation:

The density of aluminum is 2710 kg/m³.  So its volume is:

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The apparent weight is the actual weight minus the buoyant force.

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N = (9.8 m/s²) (9 kg − (1000 kg/m³) (0.0332 m³))

N = 55.7 N

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When light is incident parallel to the principal axis and then strikes a lens, the light will refract through the focal point on the opposite side of the lens.

To find the answer, we have to know about the rules followed by drawing ray-diagram.

<h3>What are the rules obeyed by light rays?</h3>
  • If the incident ray is parallel to the principal axis, the refracted ray will pass through the opposite side's focus.
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brainly.com/question/13095658

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