another physic question needs help on

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What are the total force and the absolute pressure on the bottom of a swimming pool 22.0 m by 8.5m whose uniform depth is 2.0m? What will be the pressure against the side of the pool near the bottom?

I got a force like 3665200 N, which doesn't seem correct at all. By means absolute pressure, does it mean the P in the P=F/A equation only? And how do you get pressure against the side?

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Did your calculation include atmospheric pressure?

Pressure acts in all directions so "near the bottom" would be the same as "at the bottom".

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I don't quite get what you are saying for the first part, but for the pressure at the side, it's the same way as calculating the bottom, except the height changes right?

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Well, it doesn't help to try to solve a pressure problem if you don't understand the concept of atmospheric pressure.

But the enouncing is clear: total force and absolute pressure.

So the force= P*A= 1.019716*2200*850=1906868,92 DaN without water.

Now, with water: You have to add 1.019716 DaN per ten meters of height.

As you have 2 m of height, you'll get 1.019716*1.2=1,2236592 DaN/sq. cm of absolute pressure at the bottom of the swimming pool.

Then Total Force= 1,2236592*2200*850=2288242,704 DaN

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Francis wrote:

Well, it doesn't help to try to solve a pressure problem if you don't understand the concept of atmospheric pressure.

But the enouncing is clear: total force and absolute pressure.

So the force= P*A= 1.019716*2200*850=1906868,92 DaN without water.

Now, with water: You have to add 1.019716 DaN per ten meters of height.

As you have 2 m of height, you'll get 1.019716*1.2=1,2236592 DaN/sq. cm of absolute pressure at the bottom of the swimming pool.

Then Total Force= 1,2236592*2200*850=2288242,704 DaN

OMG, you're so smart! (such a turn-on) Very Happy

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Francis wrote:

Well, it doesn't help to try to solve a pressure problem if you don't understand the concept of atmospheric pressure.

But the enouncing is clear: total force and absolute pressure.

So the force= P*A= 1.019716*2200*850=1906868,92 DaN without water.

Now, with water: You have to add 1.019716 DaN per ten meters of height.

As you have 2 m of height, you'll get 1.019716*1.2=1,2236592 DaN/sq. cm of absolute pressure at the bottom of the swimming pool.

Then Total Force= 1,2236592*2200*850=2288242,704 DaN

that's very confusing... what's 1.019716? I don't think that's the pressure of water tho... isn't P=pgh, where p is density of water, which is 1000 kg/m^3? I know that to get absolute pressure, you have to add the pressure of the water and the atmospheric pressure

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Who cares about that?

He's got Mame thrumming.

That's the main thing.

How he did it I don't know for sure.

I think they like evidence of intelligence however it is presented.

"Drop 'em. Cough. You're in.

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Mame wrote:

OMG, you're so smart! (such a turn-on)

My, my, Mame! Very Happy

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englishnewb wrote:

that's very confusing... what's 1.019716?

That's atmospheric pressure at sea level, in DaNs (DecaNewtons)..

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spendius wrote:

That's the main thing.

For sure...

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thanks a lot

btw I am doing my all my review questions to prep for my physic final, so I might need a lot of help, so please help me out Laughing

P and S waves from an earthquake travel at different speeds, and this difference helps in locating the earthquake "epicenter" (where the disturbance took place). a) Assuming typical speeds of 8.5 km/s and 5.5 km/s for P and S waves, respectively, how far away did the earthquake occur if a particular seismic station detects the arrival of these 2 types of waves 2.0 min apart? b) is one seismic station sufficient to determine the position of the epicenter? explain.

For a) I got 360 km by taking the difference, no problem there ( don't know if its right tho.. is the difference in distance between those 2 waves equals to how far the earthquake is?

for b) I got no idea...

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Well, let's have a look..

a)
You received a wave arriving at a speed of 8.5 km/s. This means that the earthquake was D (distance)=S(speed)*T(time) away or D=8.5*T.

Then, 120 s later, you received a wave that indicates the distance to the earthquake was D=5.5*(T+120).

Now, you have D=8.5*T=5.5*(T+120).

Solving that, you have D= 1870 Km.

b) You cannot know the epicentre because waves propagate in circles, you only have the general direction.

Use the triangulation method...

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I got a question about beat frequency and component frequency.

Ok, if I have 2 graphs, 1 has a beat frequency 10hz and the other 1 has a beat frequency of 50 hz, should the smaller beat frequency graph has a further component frequency compare to the other graph? But that doesn't seem to make much sense tho, since the one with 50 hz should hit the peak more often and this means the component frequency is also closer...I need some explanation haha...

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another physic question needs help on

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