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horrorfan [7]
3 years ago
9

A 3-m-high large tank is initially filled with water. The tank water surface is open to the atmosphere, and a sharp-edged 10-cm-

diameter orifice at the bottom drains to the atmosphere through a horizontal 80-m-long pipe. The total irreversible head loss of the system is determined to be 1.7 m. In order to drain the tank faster, a pump is installed near the tank exit. Determine the pump head input necessary to establish an average water velocity of 6.5 m/s when the tank is full. Disregard the effect of the kinetic energy correction factors.
Physics
1 answer:
DanielleElmas [232]3 years ago
5 0
3-m-high large tank is initially filled with water. The tank water surface is open to the atmosphere, and a sharp-edged 10-cm-diameter orifice at the bottom drains to the atmosphere through a horizontal 80-m-long pipe. If the total irreversible head loss of the system is determined to be 1.5 m, determine the initial velocity of the water from the tank. Disregard the effect of the kinetic energy correction factors.
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A student throws a rock upwards. The rock reaches a maximum height 2.4 seconds after it was released.
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Answer:

23.52 m/s

Explanation:

The following data were obtained from the question:

Time taken (t) to reach the maximum height = 2.4 s

Acceleration due to gravity (g) = 9.8 m/s²

Initial velocity (u) =..?

At the maximum height, the final velocity (v) is zero. Thus, we can obtain how fast the rock (i.e initial velocity)

was thrown as follow:

v = u – gt (since the rock is going against gravity)

0 = u – (9.8 × 2.4)

0 = u – 23.52

Collect like terms

0 + 23.52 = u

u = 23.52 m/s

Therefore, the rock was thrown at a velocity of 23.52 m/s.

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Explanation:

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8 0
3 years ago
A solid cube of aluminum (density of 2.7 g/cm³) has a volume of 0.9 cm³. how many atoms are contained in the cube?​
Reika [66]

Answer:

0.542*10^{23}\ Aluminum\ Atoms

Explanation:

We\ are\ given:\\Density\ of\ aluminum=2.7\ g/cm^3\\Volume\ of\ aluminum-cube=0.9\ cm^3\\Hence,\\As\ we\ know\ that,\\Density=\frac{Mass}{Volume}\\Mass=Density*Volume\\Hence,\ here,\\Mass\ of\ the\ solid\ iron\ cube=2.7*0.9=2.43\ g\\Now,\\We\ also\ know\ that,\\Gram\ Atomic\ mass\ of\ Aluminum = 26.98 \approx 27\ g\\Hence,\\No.\ of\ particles=\frac{Mass}{GAM}*Avagadro's Constant\\Hence,\ here\\No.\ of\ Aluminum\ atoms=\frac{2.43}{27}*6.022*10^{23} \approx 0.542*10^{23}\ Aluminum\ Atoms

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Pavlova-9 [17]

Answer:

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Explanation:

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4 years ago
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