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3 years ago

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Water is circulating through a closed system of pipes in a two floor apartment. On the first floor, the water has a gauge pressu

re of 3.70 105 Pa and a speed of 2.4 m/s. However, on the second floor, which is 3.6 m higher, the speed of the water is 3.5 m/s. The speeds are different because the pipe diameters are different. What is the gauge pressure of the water on the second floor

1 answer:

anastassius [24]3 years ago

3 0

Answer:

The value of gauge pressure at outlet = -38557.224 pascal

Explanation:

Apply Bernoulli' s Equation

$\frac{P_{1}}{9810}$ + $\frac{V_{1} ^{2}}{19.62}$ + $h_{1}$ = $\frac{P_{2}}{9810}$ + $\frac{V_{2} ^{2}}{19.62}$ + $h_{2}$ --------------(1)

Where

$P_{1}$ = Gauge pressure at inlet = 3.70105 pascal

$V_{1}$ = velocity at inlet = 2.4 $\frac{m}{sec}$

$P_{2}$ = Gauge pressure at outlet = we have to calculate

$V_{2}$ = velocity at outlet = 3.5 $\frac{m}{sec}$

$h_{2} - h_{1}$ = 3.6 m

Put all the values in equation (1) we get,

⇒ $\frac{3.70105}{9810}$ + $\frac{2.4 ^{2}}{19.62}$ = $\frac{P_{2}}{9810}$ + $\frac{3.5 ^{2}}{19.62}$ + 3.6

⇒ 0.294 = $\frac{P_{2}}{9810}$ + 0.6244 + 3.6

⇒ $\frac{P_{2}}{9810}$ = 0.294 - 0.6244 - 3.6

⇒ $\frac{P_{2}}{9810}$ = - 3.9304

⇒ $P_{2}$ = - 38557.224 pascal

This is the value of gauge pressure at outlet.

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Answers:
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Explanation:

Ooook, quick theory rushdown. if you're at a depth of h in a tank of a fluid, the pressure is the sum of the atmosferic pressure (if the tank is open on top) plus a term which is the product of acceleration of gravity - about $10 ms^-^2$ , the density of whatever you're sinking in, and the depth at which you are. In formula, $p(h) = p_0 + \rho g h$ , and the pressure is the same for every point of the tank at the same depth.

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1a. The pressure at A is - not counting atmosferic pressure - $1000 * 10 * 1 = 10^4 Pa$ , while in B is $1000*10*2 = 2*10^4 Pa$ , so it's half of it.

1b. The two points are at the same depth, so the pressure is the same - they would be even if the two cilinders weren't linked!

1c. Ditto. Same depth? same pressure!

1d. Usual equation, this time density is 800. Pressure is $800*10*2 = 1,6*10^4 Pa$ : Since the density is 4/5 of water, the pressure is also 4/5 of the one exerted by water

2a. The volume is simply the product, so $4m*3m*2m = 24m^3$

2b. Density is defined as mass over volume, so you simply multiply the volume you found earlier by the density of paraffine: $800* 24 = 1,92 *10^4kg$

2c. Weight is defined as the mass of something times the acceleration due to gravity, in our case it's $1.92 *10^4 kg * 10 ms^{-2} = 1.92 * 10^5 N$

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3. Yet again, $\rho gh$ . $1000 {kg \over m^3} * 10 {N \over kg}} * 2 m = 2* 10^4 {N\over m^2} =2*10^4 Pa$

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A 300-kg piano being held by a crane is accidentally dropped from a height of 15 meters. a. What is the speed of the piano just

FinnZ [79.3K]

Answer:

a) 17.16m/s

b) 44,145J

c) Sound the piano makes when hitting the ground, vibration of the ground, heat.

d) i) It's smaller due to the energy dissipated by the friction between air and the parachute.

ii) It stays the same, the only difference is that the dissipated energy is distributed between air resistance and the kinetic energy dissipated by the ground whent he piano hits it.

Explanation:

a)

In order to solve this problem we must start by doing a drawing of the situation, which will help us visualize the problem better. (See attached picture).

So, in this problem we can ignore air resistance so we can say that the energy is conserved, this is the total initial energy is the same as the total final energy, so we get that:

$U_{0}+K_{0}=U_{f}+K_{f}$

When the piano is released it has an initial speed of zero, so the initial kinetic energy is zero. When the piano hits the ground it will have a height of 0m, so the final potential energy is zero as well. This will simplify our equation:

$U_{0}=K_{f}$

We know that potential energy is given by the formula:

U=mgh

and kinetic energy is given by the formula:

$K=\frac{1}{2}mv^{2}$

which can be substituted in our equation:

$mgh=\frac{1}{2}mv^{2}$

we can divide both sides of the equation into the mass of the piano, so we get:

$gh=\frac{1}{2}v^{2}$

which can be solved for the final velocity which yields:

$v=\sqrt{2gh}$

we can now substitute the data provided by the problem so we get:

$v=\sqrt{2(9.81m/s^{2})(15m)}$

which yields:

v=17.16m/s

b)

Since energy is conserved, this means that the total dissipated energy will be the same as the potential energy, so we get that:

E=mgh

so

$E=(300kg)(9.81m/s^{2})(15m)$

which yields:

E=44,145J

c)

When the piano hits the ground, the kinetic energy it had will be transformed to other types of energy, mostly vibration and heat. The vibration will turn to sound due to the movement of air created by the piano itself and the ground. And heat is created by the friction between the molecules created by the vibrations and the collition itself. So some of the indicators of this release of energy could be:

-Sound

-Vibration

-Heat.

d)

i) The amount of inetic energy dissipated would decrease due to the friction between air and the parachute. Since air is resisting the movement of the piano, this will translate into a loss of energy, if we did an energy balance we would get that:

$U_{0}=K_{f}+E_{p}$

The total amount of energy is conserved but it will be distributed between the energy lost due to air resistance and the kinetic energy the piano has at the time it hits the ground.

ii) So the total amount of energy dissipated remains the same, the only difference is that it will be distributed between air resistance and the kinetic energy of the piano.

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4 years ago