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VMariaS [17]
2 years ago
10

The initial height of the water in a sealed container of diameter 100.0 cm is 5.00 m. The air pressure inside the container is 0

.850 ATM. A faucet with an opening 1.0 inch diameter is located at the bottom of the container.
Required:
a. What is the net force on the side of the container?
b. How long does it take and how much the water level will drop till water no longer comes out of the faucet?
Physics
1 answer:
katen-ka-za [31]2 years ago
4 0

Answer:

a)  F = 2.66 10⁴ N, b)   h = 1.55 m

Explanation:

For this fluid exercise we use that the pressure at the tap point is

Exterior

          P₂ = P₀ = 1.01 105 Pa

inside

         P₁ = P₀ + ρ g h

the liquid is water with a density of ρ=1000 km / m³

         P₁ = 0.85   1.01 10⁵ + 1000   9.8  5

         P₁ = 85850 + 49000

         P₁ = 1.3485 10⁵ Pa

the net force is

         ΔP = P₁- P₂

         Δp = 1.3485 10⁵ - 1.01 10⁵

         ΔP = 3.385 10⁴ Pa

Let's use the definition of pressure

         P = Fe / A

         F = P A

the area of ​​a circle is

         A = pi r² = [i d ^ 2/4

let's reduce the units to the SI system

         d = 100 cm (1 m / 100 cm) = 1 m

         F = 3.385 104 pi / 4 (1) ²

         F = 2.66 10⁴ N

b) the height for which the pressures are in equilibrium is

        P₁ = P₂

        0.85 P₀ + ρ g h = P₀

        h = \frac{P_o ( 1-0.850)}{\rho \ g}

        h = \frac{1.01 \ 10^5 ( 1 -0.85)}{1000 \ 9.8}

        h = 1.55 m

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Answer:

The value is      A   = 39315 \  m^2

Explanation:

From the question we are told that

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     The  drag coefficient is  C__{D}}  =  0.5

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     The acceleration due to gravity in Mars is  g_m  =  3.689 \  m/s^2

     

Generally the Mars  atmosphere density is mathematically represented as

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Generally the drag force on the rover and the parachute  is mathematically represented as

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=>       A   =  \frac{2 * 8374 }{ 0.5 *  0.852    *  1 ^2   }

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