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Ket [755]
3 years ago
9

g suppose a spring with spring constant of 50 N/m is hanging from the ceiling. You hang 2.0 kg mass from the spring. How far is

the spring stretched
Physics
1 answer:
adelina 88 [10]3 years ago
8 0

Answer:

The extension of the spring is 0.392 m.

Explanation:

Given;

spring constant, k = 50 N/m

mass attached to the spring, m = 2.0 kg

let the extension of the spring = x

The extension of the spring is calculated by applying Hook's law;

F = kx

mg = kx

x = \frac{mg}{k} \\\\x = \frac{2 \times 9.8}{50} \\\\x = 0.392 \ m

Therefore, the extension of the spring is 0.392 m.

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A local FM radio station broadcasts at a frequency of 97.3 MHz. Calculate the energy of the frequency at which they are broadcas
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6.45×10¯²⁶ J

Explanation:

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

Frequency (f) = 97.3 MHz

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Finally, we shall determine the energy at which the frequency is broadcasting. This can be obtained as follow:

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E = hf

E = 6.63×10¯³⁴ × 9.73×10⁷

E = 6.45×10¯²⁶ J

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