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Arturiano [62]
3 years ago
15

the volume of a water tank is 5m×4m×2m. If the tank is half filled with water. calculate the pressure exerted at the bottom of t

he tank​
Physics
1 answer:
Natali5045456 [20]3 years ago
5 0

Answer:

10⁴ Pa

Explanation:

  • Volume = 5m * 4m * 2m

When the tank is half filled ,

  • Volume = 5m * 4m * 1m = 20m³

Also we know that ,

  • Density of water = 10³ kg/ m³

\longrightarrow Density = Mass/Volume

\longrightarrow 10³ kg/m³ = m/20m³

\longrightarrow m = 2 * 10⁴ kg

So that ,

\longrightarrow Weight = mg

\longrightarrow F = 2*10⁴ × 10 N

\longrightarrow F = 2 * 10⁵ N

And ,

\longrightarrow Pressure = Force/Area

\longrightarrow P = 2 *10⁵/ 5 * 4 Pa

\longrightarrow P = 10⁴ Pa

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Answer:

The precise answer depends on the density and therefore the temperature of the water, but we can obtain a reasonable approximation by assuming that the density of the water is 1000 kilograms per cubic meter (kg/m³).

Since the depth of the water in the well is 10 m, the volume of water directly above an area A of a square meters (m²) at the bottom of the well is 10×a m³.

Since the density of the water is 1,000 kg/m³, the mass of water directly above area A is (1,000 kg/m³) × (10×a m³) = (1000×10×a kg) = 10,000×a kg.

Since g = 9.8 m/s², the force of gravity acting on the water directly above area A is (9.8 m/s²) × (10,000×a kg) = 9.8×10,000×a N (newtons) = 98,000×a N.

So the pressure of water acting on area A is (98,000×a N)/(a m²) = (98,000×a)/a N/m² = 98,000 pascals (pa). And since A could be any given area at the bottom of the well, this is the pressure at any point at the bottom of the well.

So the pressure at the bottom of the well is 98,000 pascals (or 98,000/101,325 standard atmospheres = 560/579 atmospheres ~ 0.967 standard atmospheres).

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7 0
3 years ago
A spring is 20cm long is stretched to 25cm by a load of 50N. What will be its length when stretched by 100N. assuming that the e
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Answer:

Final Length = 30 cm

Explanation:

The relationship between the force applied on a string and its stretching length, within the elastic limit, is given by Hooke's Law:

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F = Force applied

k = spring constant

Δx = change in length of spring

First, we find the spring constant of the spring. For this purpose, we have the following data:

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Δx = (100 N)/(1000 N/m)

Δx = 0.1 m = 10 cm

but,

Δx = Final Length - Initial Length

10 cm = Final Length - 20 cm

Final Length = 10 cm + 20 cm

<u>Final Length = 30 cm</u>

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