Answer:
Force = 125 [N]
Explanation:
In the attached image we can see a sketch of the lever system.
And if we make a sum of moments at the point O equal to zero (0).
In the equation showed in the image, we can determinate the force that we need
The air pressure in the pressurized tank will be 24014.88 N/m²,196.2 N/m²,2084.625 N/m².
<h3 /><h3>What is pressure?</h3>
The force applied perpendicular to the surface of an item per unit area across which that force is spread is known as pressure.
It is denoted by P. The pressure relative to the ambient pressure is known as gauge pressure.
Pressure is found as the product of the density,acceleraton due to gravity and the height.
P₁=ρ₁gh₁
P₁=13,600 kg/m³×9.81 (m/s²)×0.18 m
P₁=24014.88 N/m²
P₂=ρ₂gh₂
P₂= 1000 kg/m³×9.81 (m/s²)×00.2 m
P₂=196.2 N/m²
P₃=ρ₃gh₃
P₃=850 kg/m³×9.81 (m/s²)×0.25
P₃=2084.625 N/m²
Hence,the air pressure in the pressurized tank will be 24014.88 N/m²,196.2 N/m²,2084.625 N/m².
To learn more about the pressure refer to the link;
brainly.com/question/356585
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Answer:
The intensity of sound (I) = 3.16 x 10⁻⁶ W/m²
Explanation:
We have expression for sound intensity level (SIL),

Here we need to find the intensity of sound (I).

Substituting
L = 67 dB and I₀ = 10⁻¹² W/m² in the equation

The intensity of sound (I) = 3.16 x 10⁻⁶ W/m²
Wavelength of X-rays = 10⁻¹⁰ m
Wavelength of UV = 1000 x 10⁻¹⁰
= 10⁻⁷ m
Answer:

Explanation:
Given that:
- mass of meteoroid,

- radial distance from the center of the planet,

- mass of the planet,

<u>For gravitational potential energy we have:</u>

substituting the respective values:

