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kumpel [21]
4 years ago
10

: Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at

sea level. One container is in Denver at an altitude of about 6,000 ft and the other is in New Orleans (at sea level). The surface area of the container lid is 4 -0.0155 m. The air pressure in Denver is PD - 79000 Pa. and in New Orleans is Pro=100250 Pa. Assume the lid is weightless.
Part (a) Write an expression for the force FNo required to remove the container lid in New Orleans.
Part (b) Calculate the force Fno required to lift off the container lid in New Orleans, in newtons.
Part (c) Calculate the force Fp required to lift off the container lid in Denver, in newtons.
Part (d) is more force required to lift the lid in Denver (higher altitude, lower pressure) or New Orleans (lower altitude, higher pressure)?
Physics
1 answer:
ser-zykov [4K]4 years ago
6 0

Complete Question

Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at sea level. One container is in Denver at an altitude of about 6,000 ft and the other is in New Orleans (at sea level). The surface area of the container lid is A=0.0155 m. The air pressure in Denver is PD = 79000 Pa. and in New Orleans is PNo = 100250 Pa. Assume the lid is weightless.

Part (a) Write an expression for the force FNo required to remove the container lid in New Orleans.

Part (b) Calculate the force FNo required to lift off the container lid in New Orleans, in newtons.

Part (c) Calculate the force Fp required to lift off the container lid in Denver, in newtons.

Part (d) is more force required to lift the lid in Denver (higher altitude, lower pressure) or New Orleans (lower altitude, higher pressure)?

Answer:

a

The  expression is   F_{No} =   A [P_{No} - \frac{P_{sea}}{2}]

b

F_{No}= 7771.125 \ N

c

 F_p = 2.2*10^{6} N

d

From the value obtained we can say the that the force required to open the lid is higher at Denver

Explanation:

          The altitude of container in Denver is  d_D = 6000 \ ft = 6000 * 0.3048 = 1828.8m

           The surface area of the container lid is A = 0.0155m^2

           The altitude of container in New Orleans  is sea-level

           The air pressure in Denver is  P_D = 79000 \ Pa

            The air pressure in new Orleans is P_{ro} = 100250 \ Pa

Generally force is mathematically represented as

            F_{No} = \Delta P A

  So we are told the pressure inside is  is half the pressure the at sea level so the  the pressure acting on the container would

   The  pressure at sea level is a constant with a  value of  

               P_{sea} = 101000 Pa

So the \Delta P which is the difference in pressure within and outside the container is  

           \Delta P = P_{No} - \frac{P_{sea}}{2}

Therefore

                F_{No} =   A [P_{No} - \frac{P_{sea}}{2}]

Now substituting values

                F_{No} =   0.0155 [100250 - \frac{101000}{2}]

                       F_{No}= 7771.125 \ N

The force to remove the lid in Denver is  

           F_p = \Delta P_d A

So we are told the pressure inside is  is half the pressure the at sea level so the  the pressure acting on the container would

 The  pressure at sea level is a constant with a  value of  

               P_{sea} = 101000 Pa    

 At  sea level the air pressure in Denver is mathematically represented as

              P_D = \rho g h

     =>     g = \frac{P_D}{\rho h}      

Let height at sea level is h = 1

  The air pressure at height d_D

             P_d__{D}} = \rho gd_D

    =>     g = \frac{P_d_D}{\rho d_D}

  Equating the both

                 \frac{P_D}{\rho h}  = \frac{P_d_D}{\rho d_D}

                 P_d_D =  P_D * d_D

Substituting value  

                   P_d__{D}} = 1828.2 * 79000

                    P_d__{D}} = 1.445*10^{8} Pa

    So

              \Delta P_d  = P_{d} _D - \frac{P_{sea}}{2}

=>          \Delta P_d  = 1.445 *10^{8} - \frac{101000}{2}    

                        \Delta P_d = 1.44*10^{8}Pa

  So

               F_p = \Delta P_d A

                  = 1.44*10^8 * 0.0155

              F_p = 2.2*10^{6} N

               

                 

             

             

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