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-Dominant- [34]
3 years ago
9

Find the net downward force on the tank's flat bottom, of area 1.60 m2 , exerted by the water and air inside the tank and the ai

r outside the tank. Assume that the density of water is 1.00 g/cm3
Physics
1 answer:
Dovator [93]3 years ago
8 0

Answer:

The net downward force on the tank is 1.85\times10^{5}\ N

Explanation:

Given that,

Area = 1.60 m²

Suppose the design of a cylindrical, pressurized water tank for a future colony on Mars, where the acceleration due to gravity is 3.71 meters per second per second. The pressure at the surface of the water will be 150 K Pa , and the depth of the water will be 14.4 m . The pressure of the air in the building outside the tank will be 88.0 K Pa.

We need to calculate the net downward force on the tank

Using formula of formula

F=(P+\rho\times g\times h-P_{out})A

Where, P = pressure

g = gravity at mars

h = height

A = area

Put the value into the formula

F=(150\times10^3+1.00\times10^3\times3.71\times14.4-88.0\times10^{3})\times1.60

F=1.85\times10^{5}\ N

Hence, The net downward force on the tank is 1.85\times10^{5}\ N

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The light reactions could be viewed as analogous to a hydroelectric dam. In that case, the wall of the dam that holds back the water would be analogous to the thylakoid membrane.

Thylakoid membrane is present in cyanobacteria and chloroplasts of plants. It plays a crucial role in photosynthesis and photosystem II reactions.

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Which of the following statements is true?
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Air flows through an adiabatic turbine that is in steady operation. The air enters at 150 psia, 900oF, and 350 ft/s and leaves a
Nonamiya [84]

Answer:

1486.5\frac{Btu}{s}

Explanation:

The inlet specific volume of air is given by:

v_1=\frac{RT_1}{P_1}\\\\v_1=\frac{(0.3704\frac{psia.ft^3}{lbm.R})(1360R)}{150psia}\\\\v_1=3.358\frac{ft^3}{lbm} \ \ \ \  \ \  \ \ \...i

The mass flow rates is expressed as:

\dot m=\frac{1}{v_1}A_1V_1\\\\\dot m=\frac{1}{3.358ft^3/psia}(0.1ft^2)(350ft/s)\\\\\dot m=10.42\frac{lbm}{s}

The energy balance for the system can the be expresses in the rate form as:

E_{in}-E_{out}=\bigtriangleup \dot E=0\\\\E_{in}=E_{out}\\\\\dot m(h_1+0.5V_1^2)=\dot W_{out}+\dot m(h_2+0.5V_2^2)+Q_{out}\\\\\dot W_{out}=\dot m(h_2-h_1+0.5(V_2^2-V_1^2))=-m({cp(T_2-t_1)+0.5(V_2^2-V_1^2)})\\\\\\\dot W_{out}=-(10.42lbm/s)[(0.25\frac{Btu}{lbm.\textdegree F})(300-900)\textdegree F+0.5((700ft/s)^2-(350ft/s)^2)(\frac{1\frac{Btu}{lbm}}{25037ft^2/s^2})]\\\\\\\\=1486.5\frac{Btu}{s}

Hence, the mass flow rate of the air is 1486.5Btu/s

5 0
3 years ago
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