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drek231 [11]
3 years ago
15

Resistance depends on which three properties of a wire?

Engineering
1 answer:
kenny6666 [7]3 years ago
7 0

Answer:

2

Explanation:

From the formula R=(ro)A/l resistance depends on the length of the wire, the area of the wire(thickness) and the resistivity(ro) which depends on the material and temperature.

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(20 points) A 1 mm diameter tube is connected to the bottom of a container filled with water to a height of 2 cm from the bottom
zzz [600]

Solution :

Given :

h = 2 cm

Diameter of the tube , d = 1 mm

Diameter of the hose, D = 6 mm

Between 1 and 2, by applying Bernoulli's principle, we get

As point 1 is just below the free surface of liquid, so

$P_1=P_{atm} \text{ and} \ V_1=0$

$\frac{P_{atm}}{\rho g}+\frac{v_1^2}{2g} +h = \frac{P_2}{\rho g}$

$\frac{101.325}{1000 \times 9.81}+0.02 =\frac{P_2}{\rho g}$

$P_2 = 111.35 \ kPa$

Therefore, 111.325 kPa is the gas supply pressure required to keep the water from leaking back into the tube.

Velocity at point 2,

$V_2=\sqrt{\left(\frac{111.135}{\rho g}+0.02}\right)\times 2g$

   = 1.617 m/s

Flow of water,  $Q_2 = A_{tube} \times V_2$

                               $=\frac{\pi}{4} \times (10^{-3})^2 \times 1.617 $

                               $1.2695 \times 10^{-6} \ m^3/s$

Minimum air flow rate,

$Q_2 = Q_3 = A_{hose} \times V_3$

$V_3 = \frac{Q_2}{\frac{\pi}{4}D^2}$

$V_3 = \frac{1.2695 \times10^{-6}}{\pi\times 0.25 \times 36 \times 10^{-6}}$

    = 0.0449 m/s

b). Reynolds number in hose,

$Re = \frac{\rho V_3 D}{\mu} = \frac{V_3 D}{\nu}$

υ for water at 25 degree Celsius is $8.9 \times 10^{-7} \ m^2/s$

υ for air at 25 degree Celsius is $1.562 \times 10^{-5} \ m^2/s$

$Re_{hose}=\frac{0.0449 \times 6 \times 10^{-3}}{1.562 \times 10^{-5}}$

           = 17.25

Therefore the flow is laminar.

Reynolds number in the pipe

$Re = \frac{V_2 d}{\nu} = \frac{1.617 \times 10^{-3}}{8.9 \times 10^{-7}}$

                = 1816.85, which is less than 2000.

So the flow is laminar inside the tube.

3 0
2 years ago
g A thin-walled pressure vessel 6-cm thick originally contained a small semicircular flaw (radius 0.50-cm) located at the inner
galben [10]

This question is not complete, the complete question is;

A thin-walled pressure vessel 6-cm thick originally contained a small semicircular flaw (radius 0.50-cm) located at the inner surface and oriented normal to the hoop stress direction. Repeated pressure cycling enabled the crack to grow larger. If the fracture toughness of the material is 88 Mpam^\frac{1}{2} , the yield strength equal to 1250 MPa, and the hoop stress equal to 300 MPa, would the vessel leak before it ruptured

Answer:

length of crack is 5.585 cm

we will observe that, the length of crack (5.585 cm) is less than the vessel thickness (6 cm) Hence, vessel will not leak before it ruptures

Explanation:

Given the data in the question;

vessel thickness = 6 cm

fracture toughness k = 88 Mpam^\frac{1}{2}

yield strength = 1250 MPa

hoop stress equal = 300 MPa

we know that, the relation between fracture toughness and crack length is expressed as;

k = (1.1)(2/π)(r√(πa))  

where k is the fracture toughness, r is hoop stress and a is length of crack

so we rearrange to find  length of crack

a = 1/π[( k / 1.1(r)(2/π)]²

a = 1/π[( kπ / 1.1(r)(2)]²

so we substitute  

a = 1/π [( 88π / 1.1(300)(2/π)]²    

a = 1/π[ 0.1754596 ]

a = 0.05585 m

a = 0.05585 × 100 cm

a = 5.585 cm  

so, length of crack is 5.585 cm

we will observe that, the length of crack (5.585 cm) is less than the vessel thickness (6 cm) Hence, vessel will not leak before it ruptures

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Which of the following is likely to have the suffix "" after the domain name in its URL?
vovangra [49]

Answer:

Microsoft is the correct answer

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2 years ago
The electrical panel schedules are located on EWR Plan number ___.
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A8 is the answer because yea and because I am a teacher
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