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Aleksandr [31]
3 years ago
5

Three equal point charges, each with charge 2.00 μC , are placed at the vertices of an equilateral triangle whose sides are of l

ength 0.250 m . What is the electric potential energy U of the system? (Take as zero the potential energy of the three charges when they are infinitely far apart.) U = nothing J
Physics
1 answer:
oksano4ka [1.4K]3 years ago
4 0

Answer:

Explanation:

Energy of system of charges

= k q₁q₂ / r₁₂ + k q₁q₃ / r₁₃ + k q₃q₂ / r₃₂

q₁ , q₂ and q₃ are charges and  r₁₂ , r₁₃ , r₃₂ are densities between them

9 x 10⁹ ( 2x2 x10⁻¹²/ .25 + 2x2 x10⁻¹²/ .25 + 2x2 x10⁻¹²/ .25 )

= 9 x 10⁹  x 3 x 16 x 10⁻¹²

= 432 x 10⁻³

= .432 J .

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3 years ago
Assuming typical speeds of 8.5 km/s and 5.5 km/s for P and S waves, respectively, how far away did the earthquake occur if a par
taurus [48]

Answer:

The earthquake occurred at a distance of 1122 km

Explanation:

Given;

speed of the P wave, v₁ = 8.5 km/s

speed of the S wave, v₂ =  5.5 km/s

The distance traveled by both waves is the same and it is given as;

Δx = v₁t₁ = v₂t₂

let the time taken by the wave with greater speed = t₁

then, the time taken by the wave with smaller speed, t₂ = t₁ + 1.2 min, since it is slower.

v₁t₁ = v₂t₂

v₁t₁ = v₂(t₁ + 1.2 min)

v₁t₁ = v₂(t₁ + 72 s)

v₁t₁ = v₂t₁ + 72v₂

v₁t₁ - v₂t₁ = 72v₂

t₁(v₁ - v₂) = 72v₂

t_1 = \frac{72v_2}{v_1-v_2}\\\\t_1 =   \frac{72*5.5}{8.5-5.5}\\\\t_1 = 132 \ s

The distance traveled is given by;

Δx = v₁t₁

Δx = (8.5)(132)

Δx = 1122 km

Therefore, the earthquake occurred at a distance of 1122 km

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3 years ago
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Explanation:

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