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3 years ago

4. Water is flowing at 12m/s in a horizontal pipe under a pressure of 600kpa

Physics

1 answer:

shusha [124]3 years ago

4 0

Answer:

a. 192 m/s

b. -17,760 kPa

Explanation:

First let's write the flow rate of the liquid, using the following equation:

Q = A*v

Where Q is the flow rate, A is the cross section area of the pipe (A = pi * radius^2) and v is the speed of the liquid. The flow rate in both parts of the pipe (larger radius and smaller radius) needs to be the same, so we have:

A1*v1 = A2*v2

pi * 0.02^2 * 12 = pi * 0.005^2 * v2

v2 = 0.02^2 * 12 / 0.005^2

v2 = 192 m/s

To find the pressure of the other side, we need to use the Bernoulli equation: (600 kPa = 600000 N/m2)

P1 + d1*v1^2/2 = P2 + d1*v2^2/2

Where d1 is the density of the liquid (for water, we have d1 = 1000 kg/m3)

600000 + 1000*12^2/2 = P2 + 1000*192^2/2

P2 = 600000 + 72000 - 1000*192^2/2

P2 = -17760000 N/m2 = -17,760 kPa

The speed in the smaller part of the pipe is too high, the negative pressure in the second part means that the inicial pressure is not enough to maintain this output speed.

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2 years ago

A charge of -3.30 nC is placed at the origin of an xy-coordinate system, and a charge of 2.05 nC is placed on the y axis at y =

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Answer:

$F_{3h}=39065.298\times 10^9\ N$ attractive toward +x axis is the net horizontal force

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Explanation:

Given:

charge at origin, $Q_0=-3.35\times 10^{-6}\ C$

magnitude of second charge, $Q_2=2.05\times 10^{-6}\ C$

magnitude of third charge, $Q_3=5\times 10^{-6}\ C$

position of second charge, $(x_2,y_2)\equiv(0,4.35)\ cm$

position of third charge, $(x_3,y_3)\equiv(3.1,3.8)\ cm$

Now the distance between the charge at at origin and the second charge:

$d_2=\sqrt{(x_2-0)^2+(y_2-0)^2}$

$d_2=\sqrt{(0-0)^2+(4.35-0)^2}$

$d_2=0.0435\ m$

Now the distance between the charge at at origin and the third charge:

$d_3=\sqrt{(x_3-0)^2+(y_3-0)^2}$

$d_3=\sqrt{(3.1-0)^2+(3.8-0)^2}$

$d_3=0.04904\ m$

Now the force due to second charge:

$F_2=\frac{1}{4\pi.\epsilon_0} \times \frac{Q_0.Q_2}{d_2^2}$

$F_2=9\times 10^9\times \frac{3.3\times 2.05}{0.0435^2}$

$F_2=32175.98\times 10^9\ N$ attractive towards +y

Now the force due to third charge:

$F_3=\frac{1}{4\pi.\epsilon_0} \times \frac{Q_0.Q_3}{d_3^2}$

$F_3=9\times 10^9\times \frac{3.3\times 5}{0.04904^2}$

$F_3=61748.38\times 10^9\ N$ attractive

Now the its horizontal component:

$F_{3h}=\frac{3.1}{4.9} \times 61748.38\times 10^9$

$F_{3h}=39065.298\times 10^9\ N$ attractive toward +x axis

Now the its vertical component:

$F_{3v}=\frac{3.8}{4.9} \times 61748.38\times 10^9$

$F_{3v}=47886.49\times 10^9\ N$ upwards attractive

Now the net vertical force:

$F_v=F_{3v}+F_2$

$F_v=47886.49\times 10^9+32175.98\times 10^9$

$F_v=80062.47\times 10^9$

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Answer:

see below

Explanation:

2 Hz, means 2 complete waves are produced in one second, so

2 waves ----- 1 second

2 x 60 -------1 x 60

120 waves ------60 seconds

hope this helps,

please mark it

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