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3 years ago

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Absolute pressure in tank is P1 = 260 kPa and local ambient absolute pressure is P2 =100 kPa. If liquid density in pipe is 13600

kg/m3 , compute liquid height, h=..?.. m ? Use g =10 m/s2

1 answer:

Pavlova-9 [17]3 years ago

5 0

Answer:

1.176m

Explanation:

Local ambient pressure(P1) = 100 kPa

Absolute pressure(P2)=260kPa

Net pressure=absolute pressure-local ambient absolute pressure

Net pressure=P1(absolute pressure)-P2(local ambient absolute pressure)

Net pressure=260-100=160kPa

Pressure= ρgh

160kPa=13600*10*h

h= $\frac{160000}{136000}$

h=1.176m

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A capacitor is being charged from a battery and through a resistor of 10 kΩ. It is observed that the voltage on the capacitor ri

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3 years ago

A single conservative force F = (7.0x - 11) N, where x is in meters, acts on a particle moving along an x axis. The potential en

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Answer:

(a) 34.6429J

(b) -1.57 m

(c) 4.71 m

Explanation:

The derivative of the potential energy with respect to the position is equal to the negative of the force, so:

$-\frac{dU}{dx}=F\\ dU=-Fdx$

Then, if we integer both sides, we get:

∫dU = -∫(7x - 11)dx

$U=\frac{-7}{2}x^{2} + 11x + c$

we know that U is equal to 26 J when x is zero, so:

$U=\frac{-7}{2}x^{2} + 11x + c$

$26=\frac{-7}{2}(0)^{2} + 11(0) + c$

$26=c$

Finally, the equation for the potential energy is:

$U(x)=\frac{-7}{2}x^{2} + 11x + 26$

Therefore, the maximum positive potential energy is the energy when x is equal to 11/7. That is because the equation of U is the equation of a parable, and the vertex in a parable is given by:

$x=\frac{-b}{2a} = \frac{-11}{2(-7/2)} =\frac{11}{7}$

Where b is the number beside x and a is the number beside $x^{2}$ , Then, the value of maximum U is:

$U(11/7)=\frac{-7}{2}(11/7)^{2} + 11(11/7) + 26$

$U(11/7)=34.6429J$

On the other hand, the negative and positive values of x where the potential energy is equal to zero is calculated as:

$U(x)=\frac{-7}{2}x^{2} + 11x + 26$

$0=\frac{-7}{2}x^{2} + 11x + 26$

if we solve this using the quadratic equation, we get:

$x =\frac{-11+\sqrt{11^{2}-4(-7/2)(26)} }{2(-7/2)}=-1.5747$

$x =\frac{-11-\sqrt{11^{2}-4(-7/2)(26)} }{2(-7/2)}=4.7175$

Finally, the negative and positive values of x where the potential energy is equal to zero are -1.5747 and 4.7175 respectively.

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