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pantera1 [17]
3 years ago
15

An important dimensionless parameter in certain types of fluid flow problems is the Froude number defined as V/√g.l where V is a

velocity, the acceleration of gravity g, and a length l. Determine the value of the Froude number for v=10ft/s, g=32.2ft/s2, l=2 ft and Recalculate the Froude number using SI units for and Explain the significance of the results of these calculations.
Physics
1 answer:
prohojiy [21]3 years ago
7 0

Answer:

1.24611

Explanation:

V = Velocity = 10 ft/s

L = Length = 2 ft

g = Acceleration due to gravity = 32.2 ft/s²

Froude number is given by

Fr=\dfrac{V}{\sqrt{gL}}\\\Rightarrow Fr=\dfrac{10}{\sqrt{32.2\times 2}}\\\Rightarrow Fr=1.24611

Converting to SI units

10\ ft/s=10\times \dfrac{1}{3.281}

32.2\ ft/s^2=32.2\times \dfrac{1}{3.281}

2\ ft=2\times \dfrac{1}{3.281}

Fr=\dfrac{V}{\sqrt{gL}}\\\Rightarrow Fr=\dfrac{10\times \dfrac{1}{3.281}}{\sqrt{32.2\times \dfrac{1}{3.281}\times 2\times \dfrac{1}{3.281}}}\\\Rightarrow Fr=1.24611

The Froude number is 1.24611

The Froude number is equal. The Froude number is dimensionless as the units cancel each other. In order for this to happen the units used need to be consitent either imperial or SI.

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Refrigerant 134a enters a well-insulated nozzle at 200 lbf/in.2, 200°F, with a velocity of 120 ft/s and exits at 50 lbf/in.2 wit
riadik2000 [5.3K]

Answer:

x = 0.75801 = 75.801%

T_2 = 72..78 degree F

Explanation:

From superheated R 134 a properties table

At 200 lb/in^2 and 200 degree F

h_1 = 138.99 Btu/lbm

steady flow energy equation is givena s

h_1 + \frac{v_1^2}{2}  = h_2 + \frac{v_2^2}{2}

138.99 + \frac{120^2}{2\times 25037} = h_2 + \frac{1500^2}{2 \times 25037}

h_2 = 94.344 Btu/lbm

At 90 lb/in2 Tsat = 72.78 degree F

h_f = 35.715 Btu/lbm

hfg  = 77.345 Btu/lbm

h = hf + x hfg

94.344 = 35.715+ x \times 77.345

solving for x we get

x = 0.75801 = 75.801%

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5 0
3 years ago
Although these quantities vary from one type of cell to another, a cell can be 1.9 μm in diameter with a cell wall 60 nm thick.
DaniilM [7]

Answer: 2(10)^{-9} mg

Explanation:

We know the total diameter of the cell (assumed spherical) is:

d=1.9\mu m=1.9(10)^{-6} m

Then its total radius r=\frac{d}{2}=\frac{1.9(10)^{-6} m}{2}=9.5(10)^{-7} m

On the other hand, we know the thickness of the cell wall is r_{t}=60 nm= 60(10)^{-9} m and its density is the same as water (\rho=997 kg/m^{3}).

Since density is the relation between the mass m and the volume V:

\rho=\frac{m}{V}

The mass is: m=\rho V (1)

Now if we are talking about this cell as a thin spherical shell, its volume will be:

V=\frac{4}{3}\pi R^{3} (2)

Where  R=r-r_{w}=9.5(10)^{-7} m - 60(10)^{-9} m

Then:

V=\frac{4}{3}\pi (9.5(10)^{-7} m - 60(10)^{-9} m)^{3} (3)

V=2.952(10)^{-18} m^{3} (4)

Substituting (4) in (1):

m=(997 kg/m^{3})(2.952(10)^{-18} m^{3}) (5)

m=2.94(10)^{-15} kg (6)

Knowing 1 kg=1000 g and 1 mg=0.001 g:

m=2.94(10)^{-15} kg=2(10)^{-9} mg

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4 years ago
A car accelerates from 6.0 m/s to 18 m/s at a rate of 3.0 m/s2. How far does it travel while accelerating
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Answer:

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Based on the relative velocity of the man with respect to the boat and the dock:

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<h3>What is relative velocity?</h3>

Relative velocity is the velocity of a body relative to another body which serves as a reference point.

Relative velocity is a vector.

Considering the velocity of the man and the boat:

The relative velocity of the man with respect to the boat = 2.0 m/s

Distance covered in 4.0 seconds relative to the boat = 2.0 m/s * 4.0 s

Distance moved = 8 m

Relative velocity of the man with respect to the dock = 12 + 2 = 14 m/s

Distance covered in 4.0 seconds relative to the dock = 14.0 m/s * 4.0 s

Distance moved = 56 m

In conclusion, the relative velocity is velocity with respect to a reference point.

Learn more about relative velocity at: brainly.com/question/24337516

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