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SOVA2 [1]
4 years ago
5

A fluid in an aquifer is 23.6 m above a reference datum, the fluid pressure (in gage pressure) is 4390 n/m2 and the flow velocit

y is 7.22 x 10-4 m/s. the fluid density is 0.999 x 103 kg/m3. using the bernoulli relationship (including velocity), calculate the (a) total energy per unit mass, and (b) total energy per unit weight. report your answer in units of meters and seconds. assume that g = 9.8 m/s2.
Physics
1 answer:
Phoenix [80]4 years ago
7 0

As per Bernuolli's Theorem total energy per unit mass is given as

\frac{P}{\rho} + \frac{1}{2}v^2 + gH = E

now from above equation

P = 4390 N/m^2

\rho = 0.999 * 10^3

v = 7.22 * 10^{-4} m/s

H = 23.6 m

now by above equation

\frac{4390}{0.999*10^3} + \frac{1}{2}*(7.22*10^{-4})^2 + 9.8*23.6 = E

E = 235.7 J/kg

Part B)

Now energy per unit weight

U = \frac{E}{g}

U = \frac{235.7}{9.8}

U = 24 m

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The number of significant figures in the measurement 4.300×10^5 km are​
ICE Princess25 [194]

Answer:

6

Explanation

Any numbers in scientific notation are considered significant. For example, 4.300 x 10-4 has 4 significant figures.

Answer From Gauth Math

8 0
3 years ago
A 76 N crate is hung from a spring
lutik1710 [3]

Answer: 0.169 (3 s.f.)

Explanation:

Force = 76 N

Spring constant = 450 N/m

Extension/displacement = x

Hooke's law states that: F = kx

Therefore, 76 = 450 X x

76/450 = x

0.169 (3 s.f.) = x

4 0
3 years ago
Calculate the entropy change that occurs when 1.0kg of water at 20.00 C is mixed with 2.0kg of water at 80.00 C
SOVA2 [1]

Answer:

The change in entropy ΔS = 0.0011 kJ/(kg·K)

Explanation:

The given information are;

The mass of water at 20.0°C = 1.0 kg

The mass of water at 80.0°C = 2.0 kg

The heat content per kg of each of the mass of water is given as follows;

The heat content of the mass of water at 20.0°C = h₁ = 83.92 KJ/kg

The heat content of the mass of water at 80.0°C = h₂ = 334.949 KJ/kg

Therefore, the total heat of the the two bodies = 83.92 + 2*334.949 = 753.818 kJ/kg

The heat energy of the mixture =

1 × 4200 × (T - 20) = 2 × 4200 × (80 - T)

∴ T = 60°C

The heat content, of the water at 60° = 251.154 kJ/kg

Therefore, the heat content of water in the 3 kg of the mixture = 3 × 251.154 = 753.462

The change in entropy ΔS = ΔH/T = (753.818 - 753.462)/(60 + 273.15) = 0.0011 kJ/(kg·K).

8 0
3 years ago
The power output of a tuba is 0.35 W. At what distance is the sound
STALIN [3.7K]

The distance of the sound from the tuba is 4.82 m.

<h3>Area of the tube</h3>

The area of the tuba is calculated as follows;

I = P/A

where;

  • I is intensity of sound
  • P is power
  • A is area

A = P/I

A = 0.35 / (1.2 x 10⁻³)

A = 291.67 m²

<h3>Distance of the sound</h3>

Area = 4πr²

r = \sqrt{\frac{A}{4\pi} } \\\\r = \sqrt{\frac{291.67}{4\pi} } \\\\r = 4.82 \ m

Thus, the distance of the sound from the tuba is 4.82 m.

Learn more about intensity of sound here: brainly.com/question/4431819

8 0
3 years ago
A ball of 0.5kg slows down from 5m/s to 3m/s. Calculate the work done inthe process.
dlinn [17]

Answer:

9.8 Joules (rounded to 2 significant figures)

Explanation:

Work done (J)= Force(N) x distance changed (m)

  • Force= 9.80665 x 0.5kg
  • Force= 4.90332 Newtons

  • Distance changed= 5-3
  • distance changed= 2m/s

--> work done= 4.90332 x 2

work done= 9.8 Joules

5 0
2 years ago
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