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zavuch27 [327]
3 years ago
6

A U-tube with a cross-sectional area of 1.00 cm2 is open to the atmosphere at both ends. Water is poured into the tube until the

water rises part-way along the straight sides, and then 5.00 cm3 of oil is poured into one end. As a result, the top surface of the oil ends up 0.550 cm higher than the water surface on the other side of the U. What is the density of the oil
Physics
1 answer:
Doss [256]3 years ago
8 0

Answer:

0.89 g/cm^3 = 890 kg/m^3

Explanation:

Cross sectional area of U-tube ( A ) = 1.00 cm^2

volume of oil ( V )  = 5.00 cm^3

change between top surface = 0.550 cm

height of oil = 5 cm  ( volume / area )

height of water = 5 - 0.550 = 4.45 cm

pressure at the oil-water junction = Pressure on the second side of the U-tube at same level

Po * g * Hoil = Pw * g * Hwater

Po * 5 = 1 * 4.45

∴ Density of oil ( Po ) = 4.45 / 5  g/cm^3 = 0.89 g/cm^3

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0.599       19.168

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                 v² = v₀² - 2 a x

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                 x₁ = v₀² / 2 a₁

                         

Let's reduce speed

            v₀ = 55mph (5280 foot / 1 mile) (1h / 3600s) = 80,667 feet / s²

Let's calculate the maximum braking distance

            x₁ = 80.667² / (2 11.2)

            x₁ = 290.50 feet

The minimum braking distance

            x₂ = 80.667² / (2 19.168)

            x₂ = 169.74 feet

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            0 = v₀² - 2 a x

            v₀ = √2 a x

We calculate the speed for the two accelerations

             v₀₁ = √ (2 11.2 155)

             v₀₁ = 58.92 feet / s

       

             v₀₂ = √ (2 19.168 155)

             v₀₂ = 77.08 feet / s

To stop at the distance limit in the worst case the maximum speed must be 58.92 feet / s = 40.85 mph = 41 mph

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