Resistance depends on which three properties of a wire?

Engineering

1 answer:

kenny6666 [7]2 years ago

7 0

Answer:

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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A manometer containing a fluid with a density of 60 lbm/ft3 is attached to a tank filled with air. If the gage pressure of the a

8090 [49]

Answer:

The fluid level difference in the manometer arm = 22.56 ft.

Explanation:

Assumption: The fluid in the manometer is incompressible, that is, its density is constant.

The fluid level difference between the two arms of the manometer gives the gage pressure of the air in the tank.

And P(gage) = ρgh

ρ = density of the manometer fluid = 60 lbm/ft³

g = acceleration due to gravity = 32.2 ft/s²

ρg = 60 × 32.2 = 1932 lbm/ft²s²

ρg = 1932 lbm/ft²s² × 1lbf.s²/32.2lbm.ft = 60 lbf/ft³

h = fluid level difference between the two arms of the manometer = ?

P(gage) = 9.4 psig = 9.4 × 144 = 1353.6 lbf/ft²

1353.6 = ρg × h = 60 lbf/ft³ × h

h = 1353.6/60 = 22.56 ft

A diagrammatic representation of this setup is presented in the attached image.

Hope this helps!

5 0

3 years ago

It is desired to produce and aligned carbon fiber-epoxy matrix composite having a longitudinal tensile strength of 630 MPa. Calc

ratelena [41]

Answer:

The answer is below

Explanation:

Given that:

Diameter (D) = 0.03 mm = 0.00003 m, length (L) = 2.4 mm = 0.0024 m, longitudinal tensile strength $(\sigma_{cd})=630\ MPa = 630*10^6\ Pa$ , Fracture strength

$(\sigma_f)=5100\ MPa=5100*10^6\ Pa,fiber-matrix\ stres(\sigma_m)=17.5\ MPa=17.5*10^6\ Pa,matrix\ strength=\tau_c=17\ MPa=17 *10^6\ Pa$

a) The critical length ( $L_c$ ) is given by:

$L_c=\sigma_f*(\frac{D}{2*\tau_c} )=5100*10^6*\frac{0.00003}{2*17*10^6}=0.0045\ m=4.5\ mm$

The critical length (4.5 mm) is greater than the given length, hence th composite can be produced.

b) The volume fraction (Vf) is gotten from the formula:

$\sigma_{cd}=\frac{L*\tau_c}{D}*V_f+\sigma_m(1-V_f)\\\\V_f=\frac{\sigma_{cd}-\sigma_{m}}{\frac{L*\tau_c}{D}-\sigma_{m}} \\\\Substituting:\\\\V_f=\frac{630*10^6-17.5*10^6}{\frac{0.0024*17*10^6}{0.00003} -17.5*10^6} \\\\V_f=0.456$