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IRISSAK [1]
2 years ago
12

A hair dryer is basically a duct of constant diameter in which a few layers of electric resistors are placed. A small fan pulls

the air in and forces it through the resistors where it is heated. If the density of air is 1.20 kg/m3 at the inlet and 1.035 kg/m3 at the exit, determine the percent increase in the velocity
Physics
2 answers:
nekit [7.7K]2 years ago
5 0

Given:

density of air at inlet, \rho_{a} = 1.20 kg/m_{3}

density of air at inlet, \rho_{b} = 1.05 kg/m_{3}

Solution:

Now,

\dot{m} = \dot{m_{a}} = \dot{m_{b}}

\rho_{a} A v_{a} = \rho _{b} Av_{b}                        (1)

where

A = Area of cross section

v_{a} = velocity of air at inlet

v_{b} = velocity of air at outlet

Now, using eqn (1), we get:

\frac{v_{b}}{v_{a}} = \frac{\rho_{a}}{\rho_{b}}

\frac{v_{b}}{v_{a}} = \frac{1.20}{1.05} = 1.14

% increase in velocity = 1.14\times 100 =114%

which is 14% more

Therefore % increase in velocity is 14%

Helga [31]2 years ago
4 0

Answer:

The percent increase in the velocity is 15.9%.

Explanation:

Given that,

Density of air \rho_{in}= 1.20\ kg/m^3

Density of air \rho_{ex}=1.035\ kg/m^3

We need to calculate the percent increase in the velocity

Using continuity equation

\rho_{in}Av=\rho_{ex}Av'

\dfrac{\rho_{in}}{\rho_{ex}}=\dfrac{v'}{v}

Put the value into the formula

\dfrac{v'}{v}=\dfrac{1.20}{1.035}

\dfrac{v'}{v}=1.159

v'=1.159v

For percentage,

v'=\dfrac{1.159v-v}{v}\times100

v'=15.9\%

Hence, The percent increase in the velocity is 15.9%.

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dusya [7]

Answer:

Maximum amount of heat = 10002151.38J

Explanation:

Workdone by motor in 86.1 minutes I given by:

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The amount of heat extracted is given by:

/QL /= K/W/ = TL/W/ /(TH - TL)

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/QL/ = 390083904/39

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5 0
2 years ago
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9. A football punter attempts to kick the football so that it lands on the ground 67.0 m from where it is kicked and stays in th
vichka [17]

Answer:

Angle is 55.52°

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Put the value of v₀ from equation (1) to equation (2)

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\frac{14.99}{Cos\alpha }(Sin\alpha ) =21.828\\as\\tan\alpha =Sin\alpha /Cos\alpha \\So\\14.99tan\alpha =21.828\\tan\alpha =21.828/14.99\\\alpha =tan^{-1}(21.828/14.99) \\\alpha =55.52^{o}

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v_{o}=(\frac{14.99}{Cos\alpha } ) \\v_{o}=(\frac{14.99}{Cos(55.52) } ) \\v_{o}=26.48m/s

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