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Mkey [24]
3 years ago
6

A Venturi tube may be used as a fluid flowmeter. Suppose the device is used at a service station to measure the flow rate of gas

oline (p = 660 kg/m³) through a hose having an outlet radius of 2.80 cm. If the difference in pressure is measured to be P1 - P2 = 1.20 kPa and the radius of the inlet tube to the meter is 1.40 cm.(a) Find the speed of the gasoline as it leaves the hose.(b) Find the fluid flow rate in cubic meters per second.
Physics
1 answer:
leonid [27]3 years ago
5 0

Answer

given,

flow rate = p = 660 kg/m³

outer radius = 2.8 cm

P₁ - P₂ = 1.20 k Pa

inlet radius = 1.40 cm

using continuity equation

 A₁ v₁ = A₂ v₂

 π r₁² v₁ = π r₁² v₂

 v_1= \dfrac{r_1^2}{r_2^2} v_2

 v_1= \dfrac{1.4^2}{2.8^2} v_2

 v_1= 0.25 v_2

Applying Bernoulli's equation

 \Delta P = \dfrac{1}{2}\rho (v_2^2-v_1^2)

 \Delta P = \dfrac{1}{2}\rho (v_2^2-(0.25 v_2)^2)

 \Delta P = \dfrac{1}{2}\rho v_2^2 (1 - 0.0625)

 v_2=\sqrt{\dfrac{2\Delta P}{\rho(1 - 0.0625)}}

 v_2=\sqrt{\dfrac{2\times 1200}{660 \times(1 - 0.0625)}}

       v₂ = 1.97 m/s

b) fluid flow rate

Q = A₂ V₂

Q = π (0.014)²  x 1.97

Q = 1.21 x 10⁻³ m³/s

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Model a hydrogen atom as a three-dimensional potential well with Uo = 0 in the region 0 &lt; x a. 283 eV <br> b. 339 eV <br> c.
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This question is incomplete, the complete question is;

Model a hydrogen atom as a three-dimensional potential well with U₀ = 0 in the region 0 < x < L, 0 < y < L and 0 < z < L, and infinite otherwise, with L = 1.0 × 10⁻¹⁰ m.

Which of the following is NOT one of the lowest three energy levels of an electron in this model?

a. 283 eV

b. 339 eV

c. 113   eV  

d. 226 eV        

Answer:

the lowest three energy are; 113 eV, 225 eV, and 339 eV.

Hence Option a) 283 eV is not among the three lowest energy

Explanation:

Given the data in the question;

Three dimension cube or particle in a cubic box

the energy value is given by;

E_{nx,ny,nz = ( n_x^2 + n_y^2 + n_z^2 ) × π²h"² / 2ml²

where h" = h/2π and h is Planck's constant ( 6.626 × 10⁻³⁴ m² kg / s )

m is mass of electron ( 9.1 × 10⁻³¹ kg )

l is length of side of box ( 1.0 × 10⁻¹⁰ m )

for ground level ( n_x = n_y = n_z = 1 )

so

( n_x^2 + n_y^2 + n_z^2 ) ×  π²h"² / 2ml²

since h" = h/2π

( n_x^2 + n_y^2 + n_z^2 ) × π²h² / (2π)²2ml²

so we substitute

E_{111 = ( 1² + 1² + 1² ) × [ π²( 6.626 × 10⁻³⁴ )² ] / [ (2π)² × 2 × 9.1 × 10⁻³¹ kg × ( 1.0 × 10⁻¹⁰)² ]

E_{111 = 3 × [ (4.333188779 × 10⁻⁶⁶) / ( 7.185072 × 10⁻⁴⁹ ) ]    

E_{111 = 3 × [ 6.03082165 × 10⁻¹⁸ ]

Now, we know that electric charge = 1.602 x 10⁻¹⁹

so

E_{111 = 3 × [ (6.03082165 × 10⁻¹⁸) / (1.602 x 10⁻¹⁹) ]

E_{111 = 3 × [ 37.645578 ]

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E_{211 = ( n_x^2 + n_y^2 + n_z^2 )  × π²h² / (2π)²2ml²

we substitute

E_{211 = ( 1² + 1² + 2² ) × [ 37.645578 ]

E_{211 = 6 × [ 37.645578 ]

E_{211 = 225.87 ≈ 226 eV

E_{221 = ( n_x^2 + n_y^2 + n_z^2 )  × π²h² / (2π)²2ml²

we substitute

E_{221 = ( 2² + 2² + 1² ) × [ 37.645578 ]

E_{211 = 9 × [ 37.645578 ]

E_{211 = 338.8 ≈ 339 eV

Therefore, the lowest three energy are; 113 eV, 225 eV, and 339 eV.

Hence Option a) 283 eV is not among the three lowest energy

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