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2 years ago

Two protons (each with q = 1.60 x 10-19)

Physics

1 answer:

otez555 [7]2 years ago

5 0

Answer:

230.4 N

Explanation:

From the question given above, the following data were obtained:

Charge (q) of each protons = 1.6×10¯¹⁹ C

Distance apart (r) = 1×10¯¹⁵ m

Force (F) =?

NOTE: Electric constant (K) = 9×10⁹ Nm²/C²

The force exerted can be obtained as follow:

F = Kq₁q₂ / r²

F = 9×10⁹ × (1.6×10¯¹⁹)² / (1×10¯¹⁵)²

F = 9×10⁹ × 2.56×10¯³⁸ / 1×10¯³⁰

F = 2.304×10¯²⁸ / 1×10¯³⁰

F = 230.4 N

Therefore, the force exerted is 230.4 N

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A 0.325 g wire is stretched between two points 57.7 cm apart. The tension in the wire is 650 N. Find the frequency of first harm

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Explanation:

Given,

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we know,

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v = f λ

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