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Soloha48 [4]
2 years ago
10

In the simulation, select the pressure tab, then select the right-most of the three reservoir configurations (top left). By putt

ing weights into the small column of water and using the pressure gauge to measure the water and air pressure, estimate the cross-sectional area of the left column of water. Which of the following best represents the cross-sectional area
Physics
1 answer:
iren2701 [21]2 years ago
4 0

The volume of any rectangular solid, including a cube, is calculated mathematically as V = l × w × h

<h3>How to find a cross-sectional area?</h3>

In a given calculation, if you are given the:

  • Length
  • Width
  • Height

To find this, you would have to use the area of its base which is:

Length * Width * Height.

Furthermore, if the cross-section is parallel, then you would have to solve it thus:

Length * Width which is the area of the section from top to bottom.

Although your question is incomplete, the general idea of how to find the cross-sectional area of the volume of any rectangular solid.

Read more about cross-sectional area here:

brainly.com/question/10429397

#SPJ1

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Ba(NO₃)₂  +  Na₂SO₄  → BaSO₄ + NaNO₃

By Trial and error.

Ba(NO₃)₂  +  Na₂SO₄  → BaSO₄ + 2NaNO₃

Check:

             LHS            RHS
Ba           1                1
N             2                2
S             1                1
O      3*2 + 4 = 10         4 + 2*3 = 10
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<span>Color blindness is the failure of the red sensitive nerves in the eye that don't respond to light properly.</span>
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3 years ago
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A particle moves in a circular path of radius 0.10 m with a constant angular speed of 5 rev/s. The acceleration of the particle
Darya [45]

Answer:

2.5 m/s²

Explanation:

Acceleration: This can be defined as the rate of change of velocity.

The S.I unit of acceleration is m/s²

For circular motion, the expression for acceleration is given as,

a = ω²r ................ Equation 1

Where a = acceleration of the particle, ω = angular speed of the particle, r = radius of the circular path.

Given: ω = 5 rev/s = 31.42 rad/s, r = 0.10 m.

Substitute into equation 1

a = 5²(0.10)

a = 25(0.10)

a = 2.5 m/s²

Hence the acceleration of the particle = 2.5 m/s²

Hence, none of the option  is correct

7 0
4 years ago
For fully developed laminar pipe flow in a circular pipe, the velocity profile is u(r) = 2(1-r2 /R2 ) in m/s, where R is the inn
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Answer:

a) v_{max} = 2\ \textup{m/s}

b) v_{avg} = 1\ \textup{m/s}

c) Q = 1.256 × 10⁻³ m³/s

Explanation:

Given:

The velocity profile as:

u(r) = 2(1-\frac{r^2}{R^2} )

Now, the maximum velocity of the flow is obtained at the center of the pipe

i.e r = 0

thus,

v_{max}=u(0) = 2(1-\frac{0^2}{R^2} )

or

v_{max} = 2\ \textup{m/s}

Now,

v_{avg} = \frac{v_{max}}{2}\ \textup{m/s}

or

v_{avg} = \frac{2}}{2}\ \textup{m/s}

or

v_{avg} = 1\ \textup{m/s}

Now, the flow rate is given as:

Q = Area of cross-section of pipe × v_{avg}

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Q = \frac{\pi D^2}{4}\times v_{avg}

or

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Explanation:

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