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Mila
2 months ago
15

The water level in a tank is 20 m above the ground. a hose is connected to the bottom of the tank, and the nozzle at the end of

the hose is pointed straight up. the tank is at sea level, and the water surface is open to the atmosphere. in the line leading from the tank to the nozzle is a pump, which increases the pressure of water. if the water jet rises to a height of 30 m from the ground, determine the minimum pressure rise supplied by the pump to the water line. take the density of water to be rho = 1000 kg/m3.
Physics
1 answer:
serg [3.5K]2 months ago
5 0

Respuesta:

P_(bomba) = 98,000 Pa

Explicación:

Se nos proporciona;

h2 = 30m

h1 = 20m

Densidad; ρ = 1000 kg/m³

Primero, entendemos que la suma de las presiones en el tanque y la bomba es igual a la del boquilla,

Así, se puede expresar como;

P_(tanque)+ P_(bomba) = P_(boquilla)

Ahora, la presión se daría como;

P = ρgh

Y así,

ρgh_1 + P_(bomba) = ρgh_2

<ppor lo="" tanto="">

P_(bomba) = ρg(h_2 - h_1)

<pal sustituir="" los="" valores="" pertinentes="" obtenemos="">

P_(bomba) = 1000•9.8(30 - 20)

P_(bomba) = 98,000 Pa

</pal></ppor>
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1 month ago
9. A 2 liter bottle of Coke weighs about 2 kilograms. If that bottle were combined with an equal amount of anti-Coke, how many J
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The energy expected to be released is calculated to be 4182 Joules.

Explanation:

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1 month ago
A circular surface with a radius of 0.057 m is exposed to a uniform external electric field of magnitude 1.44 × 104 N/C. The mag
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Answer:

57.94°

Explanation:

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\Phi =E\times S\times COS\Theta

where Ф represents flux

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        θ signifies the angle between the electric field direction and the surface normal.

It is given that Ф= 78 \frac{Nm^{2}}{sec}

                          E=1.44\times 10^{4}\frac{Nm}{C}

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2 months ago
An 8.0 m, 240 N uniform ladder rests against a smooth wall. The coefficient of static friction between the ladder and the ground
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Answer:

5.7 m

Explanation:

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AB = the position of the ladder's center of mass = (0.5) L = (0.5) 8 = 4 m

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F = force exerted by the wall on the ladder

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By applying force equilibrium in the vertical direction

N = F_{g} + W

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f = static friction force on the ladder

Static friction force can be expressed as

f = μ N

f = (0.55) (950)

f = 522.5 N

The equation for force along the horizontal axis reads

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F = 522.5 N

using torque equilibrium around point A

F Sin50 (AD) = W Cos50 (AB) + (F_{g} Cos50 (AC))

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Answer:

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