Exploring Venus. The surface pressure on Venus is 92 atm, and the acceleration due to gravity there is 0.894g. In a future exploratory mission, an upright cylindrical tank of benzene is sealed at the top but still pressurized at 92 atm just above the benzene. The tank has a diameter of 1.72 m, and the benzene column is 11.50 m tall. Ignore any effects due to the very high temperature on Venus. What force does the Venusian atmosphere exert on the outside surface of the bottom of the tank?

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Hi everyone. In this particular practice problem, we are asked to find the force experience at the bottom of the container due to Jupiter's atmosphere where it will have an atmospheric pressure of 9. 80 m and gravitational acceleration of 2.63 G, we will have a sealed circular container containing soda and compressed air at a pressure of 9.87 80 m above the soda and taken to Jupiter for an experimental study. The soda column is 2 25.4 m high and the container is actually having a diameter of 1.6 m. Okay. So we are asked to actually determine the force that the bottom of the container is experiencing because of Jupiter's atmosphere. So we want to recall that pressure can actually be calculated by um defining force with its cross sectional area where the pressure is or where the force is applied to and using and rearranging this formula. We can find that the force can be calculated by multiplying the pressure with the cross sectional area. And then because of the atmospheric effect and everything else, we can calculate pressure by P equals to P A T M plus H rho G just like. So, so we are required to actually determine the force on an area due to the pressure and we will determine the area and also the pressure and then solve for F from this, from this formula right here. So we want to actually use the pressure in Pascal's because that's the S I unit. So one pascal is going to be one Newton per meter squared and Newton per meter squared are both as I. So first we note that the atmospheric pressure in Jupiter is 9.87 80 M. And we want to actually um conferred this to Pascal's. So the way we want to do that is 1 80 M will correspond to 101.3 K P A or kilo Pascal's And this is 180 M and one kg Pascal's will equals to 1000 Pascal's over one kg, Pascal's just like. So, so we can cross out all the other units and the pressure pressure will then be 9. times 10 to the power of five Pascal just like that. So next, we want to actually calculate the area of where the pressure is applied to or the forces applying to. So the area is going to equals two because it is a circular container is just essentially pi R squared. And we are given the diameter of 1.6 m. So essentially a diameter is going to be twice the radius. So this is going to be D over two squared. So this is going to be pie multiplied by 1.6 m over two squared. And that will corresponds to a cross sectional area of 2. m squared just like. So okay, finally, we can actually calculate our force here. So the force will equals to P multiplied by A. In this case, we already know the P which is 9.998 times 10 to the power of five Pascal's And the area is 2. m squared. And that will correspond to a force of 2.01 times 10 to the power of six Newton. And that will be the answer to this particular practice problem. 2.01 time standard part of six Newton will correspond to option b. So option B is going to be the answer to this problem and that will be all for this video. If you guys have any sort of confusion, please make sure to check out our other lesson videos on our similar topics and that will be all for this one. Thank you.

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Key Concepts

Pressure

Force

Area of a Circle