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3 years ago

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An incandescent light bulb can be regarded as a resistor. If its power output is 100W, calculate the resistance of the light bul

b and the current drawn by the light bulb, assuming the voltage of the power source is 110V.

1 answer:

stira [4]3 years ago

8 0

Answer: $121\ \Omega$

$0.909\ A$

Explanation:

Given

Power $P=100\ W$

Voltage applied $V=110\ V$

Resistance of the bulb is given by

$P=\frac{V^2}{R}$

$100=\frac{110^2}{R}$

$R=\frac{12100}{100}$

$R=121\ \Omega$

Current drawn by the Power source is given by

$P=V\cdot I$

$I=\frac{P}{V}$

$I=\frac{100}{110}$

$I=0.909\ A$

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Air is compressed isothermally from 13 psia and 55°F to 80 psia in a reversible steady-flow device. Calculate the work required,

solong [7]

Answer:64.10 Btu/lbm

Explanation:

Work done in an isothermally compressed steady flow device is expressed as

Work done = P₁V₁ In { P₁/ P₂}

Work done=RT In { P₁/ P₂}

where P₁=13 psia

P₂= 80 psia

Temperature =°F Temperature is convert to °R

T(°R) = T(°F) + 459.67

T(°R) = 55°F+ 459.67

=514.67T(°R)

According to the properties of molar gas, gas constant and critical properties table, R which s the gas constant of air is given as 0.06855 Btu/lbm

Work = RT In { P₁/ P₂}

0.06855 x 514.67 In { 13/ 80}

=0.06855 x 514.67 In {0.1625}

= 0.06855 x 514.67 x -1.817

=- 64.10Btu/lbm

The required work therefore for this isothermal compression is 64.10 Btu/lbm

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3 years ago

Calculate the tensile modulus of elasticity for a laminated composite consisting of 62 percent by volume of unidirectional carbo

kompoz [17]

Answer:

4.30 gp

Explanation:

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3 years ago

Type the correct answer in the box. Spell all words correctly.

sesenic [268]

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3 years ago

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Water flows through a converging pipe at a mass flow rate of 25 kg/s. If the inside diameter of the pipes sections are 7.0 cm an

ser-zykov [4K]

Answer:

volumetric flow rate = $0.0251 m^3/s$

Velocity in pipe section 1 = $6.513m/s$

velocity in pipe section 2 = 12.79 m/s

Explanation:

We can obtain the volume flow rate from the mass flow rate by utilizing the fact that the fluid has the same density when measuring the mass flow rate and the volumetric flow rates.

The density of water is = 997 kg/m³

density = mass/ volume

since we are given the mass, therefore, the volume will be mass/density

25/997 = $0.0251 m^3/s$

volumetric flow rate = $0.0251 m^3/s$

Average velocity calculations:

<em>Pipe section A:</em>

cross-sectional area =

$\pi \times d^2\\=\pi \times 0.07^2 = 3.85\times10^{-3}m^2$

mass flow rate = density X cross-sectional area X velocity

velocity = mass flow rate /(density X cross-sectional area)

$velocity = 25/(997 \times 3.85\times10^{-3}) = 6.513m/s$

<em>Pipe section B:</em>

cross-sectional area =

$\pi \times d^2\\=\pi \times 0.05^2= 1.96\times10^{-3}m^2$

mass flow rate = density X cross-sectional area X velocity

velocity = mass flow rate /(density X cross-sectional area)

$velocity = 25/(997 \times 1.96\times10^{-3}) = 12.79m/s$

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3 years ago

A vacuum pump is used to drain a basement of 20 °C water (with a density of 998 kg/m3 ). The vapor pressure of water at this tem

lord [1]

Answer:

The maximum theoretical height that the pump can be placed above liquid level is $\Delta h=9.975\,m$

Explanation:

To pump the water, we need to avoid cavitation. Cavitation is a phenomenon in which liquid experiences a phase transition into the vapour phase because pressure drops below the liquid's vapour pressure at that temperature. As a liquid is pumped upwards, it's pressure drops. to see why, let's look at Bernoulli's equation:

$\frac{\Delta P}{\rho}+g\, \Delta h +\frac{1}{2} \Delta v^2 =0$

( $\rho$ stands here for density, $h$ for height)

Now, we are assuming that there aren't friction losses here. If we assume further that the fluid is pumped out at a very small rate, the velocity term would be negligible, and we get:

$\frac{\Delta P}{\rho}+g\, \Delta h =0$

$\Delta P= -g\, \rho\, \Delta h$

This means that pressure drop is proportional to the suction lift's height.

We want the pressure drop to be small enough for the fluid's pressure to be always above vapour pressure, in the extreme the fluid's pressure will be almost equal to vapour pressure.

That means:

$\Delta P = 2.34\,kPa- 100 \,kPa = -97.66 \, kPa\\$

We insert that into our last equation and get:

$\frac{ \Delta P}{ -g\, \rho\,}= \Delta h\\\Delta h=\frac{97.66 \, kPa}{998 kg/m^3 \, \, 9.81 m/s^2} \\\Delta h=9.975\,m$

And that is the absolute highest height that the pump could bear. This, assuming that there isn't friction on the suction pipe's walls, in reality the height might be much less, depending on the system's pipes and pump.

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3 years ago