QUESTION 9 What is the total force on the bottom of a...

QUESTION 9 What is the total force on the bottom of a swimming pool 27.4 m by 29 m whose uniform depth is 3.9 m? g = 9.8 m/s^2 density of water = 1000 kg/m^3 Atmospheric pressure = 1.013 x 10^5 Pa Give the answerr in scientific notation. For example If your answer is 1.2 x 10^5 write 1.2e5

Transcript

00:01 So here on this problem we are given that the dimension of the swimming pool at the bottom is this one is 29 meter and this dimension we are given as 27 .4 meter now the depth of the swimming pool is given us 3 .9 meter now the atmospheric pressure p not acting on the swimming pool is 1 .013 times 10 to the power 5 pascal now we have to calculate the total force exertion by the water at the bottom of the swimming pool that means at this point now let me first calculate the pressure acting at this point and after calculating the pressure we can easily calculate force because we know that force is equal to pressure times the area so let me calculate the pressure so pressure is equal to now here the pressure is p -0 so now when we are moving in the downward direction the pressure will be going to increase by the magnitude of row g and h so pressure at this point would be equals to p .0 plus row g h so p note that is atmospheric pressure we are given us 1 .013 times 10 to the power 5 pascal means newton per meter square plus the density so density of the water we are given as 1 ,000 kg per meter cube multiplied with g that is the acceleration due to gravity and its value is 9 .8 meter per second squared and the height edge this height edge would mean this depth which we are given as 3 .9 meter now can simplify this and we'll get our pressure at this point at the bottom to be equals to 1 .395 times 10 to the power 5 newton per meter square so this is the pressure acting at the bottom at every point at every this point the pressure would be acting this much of amount now from here we can easily calculate the force so force we know it is equals to pressure times the cosectional area so pressure is 1 .395 times 10 to the power 5 newton per meter square times the cosection area would mean this dimension multiplied with this dimension so 27 .4 meter multiplied with 3 .9 meter now you can simplify this to get our result and we'll get our answer to be nearly 11 .0 .8 9 times 10 to the power 6 newton.

02:47 So this is the answer for the given problem...

Question

QUESTION 9 What is the total force on the bottom of a swimming pool 27.4 m by 29 m whose uniform depth is 3.9 m? g = 9.8 m/s^2 density of water = 1000 kg/m^3 Atmospheric pressure = 1.013 x 10^5 Pa Give the answerr in scientific notation. For example If your answer is 1.2 x 10^5 write 1.2e5

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