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Nookie1986 [14]
3 years 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:
Damm [24]3 years ago
5 0

Answer:

P_(pump) = 98,000 Pa

Explanation:

We are given;

h2 = 30m

h1 = 20m

Density; ρ = 1000 kg/m³

First of all, we know that the sum of the pressures in the tank and the pump is equal to that of the Nozzle,

Thus, it can be expressed as;

P_(tank)+ P_(pump) = P_(nozzle)

Now, the pressure would be given by;

P = ρgh

So,

ρgh_1 + P_(pump) = ρgh_2

Thus,

P_(pump) = ρg(h_2 - h_1)

Plugging in the relevant values to obtain;

P_(pump) = 1000•9.8(30 - 20)

P_(pump) = 98,000 Pa

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025 10.0 points
mamaluj [8]

Answer:

m_b=278.73\ kg

Explanation:

<u>Center of Gravity </u>

It refers to a point where all the forces of gravity of a body make a zero total torque. To find the solution we use the fact that the net force acting on the system boat-man is in every moment equal to zero. It's assured by the first Newton’s law, the center of gravity is at rest or in uniform motion in both moments. From an external viewer's point of view, the center of gravity remains unchanged. The formula to compute it is shown below

\displaystyle x_c=\frac{\sum x_im_i}{\sum m_i}

Originally, the man sits on the stern of the boat. His weight is applied at a distance xm=4.9 m from the pier (assumed as x=0). The boat is assumed to have a uniformly distributed mass applied at its center, i.e. at xb = 4.9 / 2 = 2.45 m. The center of gravity is located originally at

\displaystyle x_c=\frac{(4.9)(90.4)+(2.45)(m_b)}{90.4+m_b}

\displaystyle x_c=\frac{442.96+2.45m_b}{90.4+m_b}

When the man walks to the prow, the boat moves x = 1.2 m from the pier, so its center is located at a distance  

x_b=1.2+2.45=3.65\ m

The man is located at  

x_m=1.2\ m

The center of gravity is computed now as

\displaystyle x_c=\frac{(1.2)(90.4)+(3.65)(m_b)}{90.4+m_b}

\displaystyle x_c=\frac{108.48+3.65m_b}{90.4+m_b}

Both centers of gravity are equal, thus

\displaystyle \frac{442.96+2.45m_b}{90.4+m_b}=\frac{108.48+3.65m_b}{90.4+m_b}

Simplifying

442.96+2.45m_b=108.48+3.65m_b

Rearranging

1.2m_b=334.48

Thus

\boxed{m_b=278.73\ kg}

6 0
3 years ago
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