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A deep water progressive wave has a height of 4 meters and a period of 8 seconds.

1.What is the wavelength?
2.What is the max horizontal acceleration on the mean water level?
3.What is the total average energy induced by the wave. Density of seawater id 1025kg/m³?

User Hal
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Final answer:

The wavelength of the deep water progressive wave is approximately 25.08 meters. The maximum horizontal acceleration on the mean water level is approximately 0.79 m/s². The total average energy induced by the wave is approximately 79603.3 Joules.

Step-by-step explanation:

To find the wavelength of a deep water progressive wave, we can use the formula: wavelength (λ) = (g × T²) / (2π), where g is the acceleration due to gravity (9.8 m/s²) and T is the period of the wave (8 s). Plugging in the values, we get: λ = (9.8 × 8²) / (2π) ≈ 25.08 meters.

The maximum horizontal acceleration on the mean water level can be determined using the formula: a = (2π × H) / T², where H is the height of the wave (4 m) and T is the period of the wave (8 s). Substituting the values, we get: a = (2π × 4) / 8² ≈ 0.79 m/s².

The total average energy induced by the wave can be calculated using the formula: E = (1/16) × ρ × g × H² × λ, where ρ is the density of the seawater (1025 kg/m³), g is the acceleration due to gravity (9.8 m/s²), H is the height of the wave (4 m), and λ is the wavelength of the wave (25.08 m). Plugging in the values, we get: E = (1/16) × 1025 × 9.8 × 4² × 25.08 ≈ 79603.3 Joules.

User Avenmia
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