a-swimming-pool-is-600-mathrmm-wide-and-150-mathrmm-long-the-bottom-has-a-constant-slope-such-that-3

Name: a swimming pool is 600 mathrmm wide and 150 mathrmm long the bottom has a constant slope such that 3
Uploaded: 2021-12-10
Duration: 2 min 49 s
Channel: Numerade
Description: A swimming pool is $6.00 \mathrm{m}$ wide and $15.0 \mathrm{m}$ long. The bottom has a constant slope such that the water is $1.00 \mathrm{m}$ deep at one end and $2.00 \mathrm{m}$ deep at the other end. Find the force of the water on one of the sides of the pool.

Question

A swimming pool is $6.00 \mathrm{m}$ wide and $15.0 \mathrm{m}$ long. The bottom has a constant slope such that the water is $1.00 \mathrm{m}$ deep at one end and $2.00 \mathrm{m}$ deep at the other end. Find the force of the water on one of the sides of the pool.

Stanley Enemuo · Accepted Answer

the length of the cable that is described. This equation is Y. Is equal to Your .04 Times exit in 1/2. I mean excellent. We have From X is equal to 0 2. X. is equal to 100 ft. So the arc line give me defined us as equal to the integral of the square root of one plus. Dy was the X. Square. Yes, but the limits being A and B. So now we have to take the french show of D. Y. D. I. Again a part of chain rule today. So dy over dx is equal to zero point 06 X. & 1/2. Now we could now use our clan safety tuning villa for the X. Into this equation. You can now Valerie or integral. So we have the integral of 0: 100. You have one close zero point 0086 X. D. X. And that gives us 108 points part two. And that is the length of our cable.

Surjit Tewari · Answer

question What 12 First of all, we will find the hydrostatic pressure added the top 2 m. This pressure is equal to density of water role in the G into the depth edge. Stop shooting the values Ro is 1000 KD permitted you g value 9.8 m per second disc wired and that there&#x27;s 2 m so we will get the hydrostatic pressure 19 600 Paschal. Now, in the first part of the question, the force at the bottom is equal to the hydrostatic pressure into the area off the bottom. So pressure is 19 600 and area off bottom is five team eaten into 10 m. So this will give us the force five points it Yeah, into 10 to the power six. No done now in part way, the force at each end is equal to the average pressure on each end. Which is he by two into the area off each end. So substituting the values Russia is 19 600 divided by two into area just to into Then we tell So this will give us the pressure. A teaching 1.96 into 10 to the power five Newton. Now, in the last part of the question, the force at each side is equal to the average pressure, which is 19600 They were dead by two into the area off each side, which is 30 m into 2 m. Solving this, we will get the force on each side, 5.88 into 10 to the power five Newton.

Mayukh Banik · Answer

The pressure at the bottom of the pool is given by the density off water G times hide of the our depth of the water. So that comes out to the pressure of the density of the water is 1000. Gravity is 9.8, and, uh, tank is filled up upto eight equals two meters. So that comes out to the 1.96 time stand to the fore basket. So force at the bottom becomes pressure at the bottom times any of the bottom. That is 1.96 times 10 to the four Paschal times, 30 meter by 10 meter, and this comes out to be 5.88 time. Stand to the six Newton. Now, remember that if this is the swimming pool, this side is 30 and that side is 10 on the Heidi&#x27;s, too. So we have got to different areas. Area one and area, too. Now for area one. Our area, too, for area one or area to the effective hide is the half of the height it was watt meter, so force inside one is given by pressure on side one times Area one. It was pressure one of these lo g h over to times area Juan. So that comes out to be 1.96 times 10 to the four, 1.96 times 10 to the four times line, born eight times one times, then times two new done. And for the second surface, it will come out to be similar as 1.96 times 10 to the four times 9.8 times one. This is still the pressure, but the idea now is 30 by to new time solving for the numbers. If one an F two comes out to be the 1st 1 comes out to be 1.96 times stand to the five Newton and the 2nd 1 comes out to be 5.88 10 to the five nudo.

Averell Hause · Answer

so here for a party, we can say the weight of the water would be equaling the density times G times be and so the mass would be the density times of volume. And then, uh, the mass times acceleration due to gravity would give us the weight and so this would be equal to the density of water. 1000 kilograms per cubic meter multiplied by 9.80 meters per second squared. This would be multiplied by the volume 5.0 meters multiplied by 4.0 meters times 3.0 Brother being the same multiplied 3.0 meters and this would equal 5.9 times 10 to the fifth Newton&#x27;s. This would be the weight of the water. Now for part B. Integration gives the expected result that the force is what it would be if the pressure were uniform and equal to the pressure at the midpoint. So essentially we could find the force. This would be equaling the density times acceleration due to gravity times. The area multiplied by D over too. And so we find that this is equaling 1000 kilograms per cubic meter multiplied by 9.8 zero meters per second. Squared multiplied. Rather, we can say for 0.0 meters multiplied by 3.0 meters multiplied by D over too 1.50 meters. And we find that the force is equaling 1.76 times 10 to the fifth Newton&#x27;s. This would be your answer for part B. That is the end of the solution. Thank you for

Averell Hause · Answer

we know that the water exerts a small force a differential force Equalling the pressure most applied by the infinite testily small, uh, cross sectional area. This would be the pressure multiplied by a constant with multiply by, of course, a changing. Why her changing height? In this case, we can say that the pressure p would be equaling row G D minus. Why? Analyzing the dam? And so the pressure, of course. At depth D minus. Why now we can say that all together we can say the DF then be equaling two row g w d minus y d y. And so we can say that then the total force would be equaling two the integral and essentially all slices of the dam because we&#x27;re taking the differential force for each slice of essentially each, um, slice, horizontal slice and we&#x27;re adding them all up. And this would be of the differential force, that infinite testimony, small force. And so this would be, of course, integrated from zero to D for DF. This would then be equaling two row G w These air constants multiplied by the integral from zero to D uh D minus y de y moving on. We find that then the total force is that equaling row G w multiply by Why d minus. Why squared over to simply taking the integral evaluated at zero and that d And so this is going to be giving us the total force equaling 1/2 multiplied by row G W d squared. And so I would be the formula for part B. We simply have to plug in. So the total force would then be equal to 1/2 multiplied by 1000 kilograms per cubic meter. Density of water multiplied by acceleration due to gravity, 9.80 meters per second squared, multiplied by the with of 100 meters multiplied by the depth of 60 meters Quantity squared. And we find that then the total force is gonna be equaling 1.76 times 10 raised to the ninth Power Newtons. That is the end of the solution. Thank you for watching

Bobby Barnes · Answer

So we have this rectangular swimming pool and it is 100 feet by 1 50 p. And it contains 100 200,000 gallons of water. We want to determine how deep it is, what looks think about this. It may be drawing a picture will help. So this is our pool. So it&#x27;s 100 feet by 150 feet, and it is some about deep in. We don&#x27;t know how deep it is yet, So we&#x27;ll just call this our depth D. Well, you know, the value of a rectangular prison, which is what we have here is length, times, width, times, depth for enormous, a height book, same thing. And we want to solve her depth so we could just go ahead and divide each side by L never W. And that&#x27;s going to give us that De is he going to be over L W. And actually, it&#x27;s this over a little bit because I&#x27;ll need to do some conversions. Well, that Elle Times W is what we have right here. So just go and figure out what that is. There&#x27;s gonna be one 15,000 feet squared so square feet and Now our volume is a gallons, but all of our other units has to deal with feet. So we want to convert this to beat somehow. And in the previous problem, they tell us that one cubic foot has about 7.5 gallons of water. So we have one cubic what? And we divide by gallons. So the gallon units Cat&#x27;s out with each other and I&#x27;m just gonna leave this as 200,000 over 7.5 cubic part. So we have all this. Now let&#x27;s plug everything into our equation. So D is going to be equal to Or volume is 200,000 divided by 7.5 and now we&#x27;re going to knew L Times W which we found to be 15,000. So our units in the numerator were cubed and the denominator square, but in a Y road inches feet, Sophie, cute and Pete&#x27;s Square. So those units should just cancel down too beat and that we were to divide everything. Let&#x27;s see what we would get. So too 100. How&#x27;s it divided by 7.5 divided by 1500 this simplifies down to 16/9 beat, which is approximately equal to 1.78 So this here is going to be how deep are Polish

Problem

Find the force on the lower half of the wall at t…

Problem 25 Easy Difficulty

A swimming pool is $6.00 \mathrm{m}$ wide and $15.0 \mathrm{m}$ long. The bottom has a constant slope such that the water is $1.00 \mathrm{m}$ deep at one end and $2.00 \mathrm{m}$ deep at the other end. Find the force of the water on one of the sides of the pool.

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