A uniform beam resting on two pivots has a length L = 6.00 m and weight M = 220 lbs. The pivot under the left end exerts a norma

Response:

a) , b)

Clarification:

a) The absolute pressure at a depth of 27.5 meters is:

b) The force applied by the water is:

V = Volume of gas sample = 1.00 L = 0.001 m³T = temperature of gas = 25.0 °C = 25 + 273 = 298 K P = pressure = 1.00 atm = 101325 Pa n = number of moles of gas using ideal gas law:PV = n RT101325 (0.001) = n (8.314) (298)n = 0.041 n₁ = moles of heliumn₂ = moles of neonm₁ = mass of helium = n₁ (4) = 4 n₁m₂ = mass of neon = n₂ (20.2) = 20.2 n₂given that:m₁ = m₂4 n₁ = 20.2 n₂n₁ = 5.05 n₂also n₁ + n₂ = n5.05 n₂ + n₂ = 0.041n₂ = 0.0068mole fraction of neon is mole fraction = n₂ /n = 0.0068/0.041 = 0.166P₂ = partial pressure of neon =(mole fraction) P P₂ = (0.166) (1)P₂ = 0.166 atm

1) The buoyant force acting on an object submerged in a fluid can be described as:

where indicates the fluid's density, represents the volume of the fluid displaced, and signifies the gravitational acceleration.

2) To determine the volume of the displaced fluid, we note that the titanium object is entirely submerged in the fluid (air), thus this volume matches the volume of 1 Kg of titanium, which has a density of . Using the correlation between density, volume, and mass, we derive


3) We can now revisit the equation in step 1) to compute the buoyant force. Given that the air density is , this provides us with


4) The weight of 1 Kg of titanium is:

Therefore, the buoyant force is negligible when compared to the weight.