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3 years ago

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A metal conduit will be used as a pathway for wiring through a concrete block. The conduit is a 4 foot long rod with an outer di

ameter of 2.2 inches and an inner diameter of 2 inches. What is the volume of metal in the conduit in cubic centimeters? ...?

2 answers:

AlekseyPX3 years ago

7 0

Answer: $510. 12 cm^3$

The volume of a cylindrical conduit, $V=\pi (r_2^2-r_1^2) h$

where $r_1$ is the inner radius, $r_2$ is the outer radius, h is the length.

First of all, we would covert each parameter into centimeters.

length of the conduit, $h=4 ft=4\times 30.48 cm=121.92 cm$

outer radius, $r_1=2.2/2 in =1.1 in= 1.1\times 2.54 cm=2.79 cm$

inner radius, $r_1=2/2 in =1 in= 2.54 cm$

Insert the values in the volume formula,

$V=3.14 \times (2.79^2-2.54^2) \times 121.92 cm^3=510. 12 cm^3$

Hence, the volume of metal in the conduit is 510. 12 cubic centimeters.

V125BC [204]3 years ago

5 0

We assume this conduit is in a cylindrical form. So that the volume can be calculated by the expression given as:

V=πr²h

However, we the conduit has a hollow space and this should not be included in the volume. We calculate as follows:

V(outer cylinder) = πr²h = π(2.2/2)²(4)(12) = 182.46 in³
V(inner cylinder ) = πr²h = π(2/2)²(4)(12) = 48π in³
V = 182.46 in³ - 48π in³ = 31.66 in³ ---->ANSWER

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<u>Answer:</u> The number of photons are $3.7\times 10^8$

<u>Explanation:</u>

We are given:

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$E=\frac{hc}{\lambda}$

where,

h = Planck's constant = $6.625\times 10^{-34}J.s$

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Putting values in above equation, we get:

$E=\frac{6.625\times 10^{-34}J.s\times 3\times 10^8m/s}{0.122m}\\\\E=1.63\times 10^{-24}J$

Now, calculating the energy of the photon with 88.3 % efficiency, we get:

$E=1.63\times 10^{-24}\times \frac{88.3}{100}=1.44\times 10^{-24}J$

To calculate the mass of water, we use the equation:

$Density=\frac{Mass}{Volume}$

Density of water = 1 g/mL

Volume of water = 165 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{165mL}\\\\\text{Mass of water}=165g$

To calculate the amount of energy of photons to raise the temperature from 23°C to 100°C, we use the equation:

$q=mc\Delta T$

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c = specific heat capacity of water = 4.184 J/g.°C

$\Delta T$ = change in temperature = $T_2-T_1=100^oC-23^oC=77^oC$

Putting values in above equation, we get:

$q=165g\times 4.184J/g.^oC\times 77^oC\\\\q=53157.72J$

This energy is the amount of energy for 'n' number of photons.

To calculate the number of photons, we divide the total energy by energy of one photon, we get:

$n=\frac{q}{E}$

q = 53127.72 J

E = $1.44\times 10^{-24}J$

Putting values in above equation, we get:

$n=\frac{53157.72J}{1.44\times 10^{-24}J}=3.7\times 10^{28}$

Hence, the number of photons are $3.7\times 10^8$

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Putting the given values above and we will calculate the value of length as follows.

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