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ahrayia [7]
3 years ago
15

A fluid moves through a tube of length 1 meter and radius r=0.002±0.0002 meters under a pressure p=4⋅105±1750 pascals, at a rate

v=0.5⋅10−9 m3 per unit time. Use differentials to estimate the maximum error in the viscosity η given by
Physics
1 answer:
yaroslaw [1]3 years ago
4 0

Answer:

The  maximum error is  \Delta  \eta  = 2032.9

Explanation:

From the question we are told that

     The length  is  l  =  1\ m

      The radius is  r =  0.002 \pm  0.0002 \ m

        The pressure is  P  =  4 *10^{5} \ \pm 1750

        The  rate  is  v =  0.5*10^{-9} \ m^3 /t

       The viscosity is  \eta  =  \frac{\pi}{8} * \frac{P *  r^4}{v}

The error in the viscosity is mathematically represented  as

       \Delta  \eta  = | \frac{\delta \eta}{\delta P}| *  \Delta  P   +    |\frac{\delta \eta}{\delta r} |*  \Delta  r +  |\frac{\delta \eta}{\delta v} |*  \Delta  v

   Where  \frac{\delta \eta }{\delta P} =  \frac{\pi}{8} *  \frac{r^4}{v}

and         \frac{\delta \eta }{\delta r} =  \frac{\pi}{8} *  \frac{4* Pr^3}{v}

and          \frac{\delta \eta }{\delta v} =  - \frac{\pi}{8} *  \frac{Pr^4}{v^2}

So  

             \Delta  \eta  = \frac{\pi}{8} [ |\frac{r^4}{v}  | *  \Delta  P   +    | \frac{4 *  P * r^3}{v}  |*  \Delta  r +  |-\frac{P* r^4}{v^2}  |*  \Delta  v]

substituting values

            \Delta  \eta  = \frac{\pi}{8} [ |\frac{(0.002)^4}{0.5*10^{-9}}  | *  1750   +    | \frac{4 *  4 *10^{5} * (0.002)^3}{0.5*10^{-9}}  |*  0.0002 +  |-\frac{ 4*10^{5}* (0.002)^4}{(0.5*10^{-9})^2}  |*  0 ]

  \Delta  \eta  = \frac{\pi}{8} [56  +  5120 ]

   \Delta  \eta  = 647 \pi

    \Delta  \eta  = 2032.9

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The question is incomplete. The complete question is :

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The normal force represented by N is equal to the downward force, $F_d$ which is equal in magnitude but it is opposite in direction.

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Answer:

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Explanation:

The apparent weight of a body is the weight due to the gravitational attraction minus the thrust due to the fluid where it will be found.

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Answer:

a) a=33.73mm/s^{2}

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In this case, we know the centripetal acceleration is given by the following formula:

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we know the period of rotation of the earth is about 24 hours, so:

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a_{c}=33.73mm/s^{2}

b)

In order to answer part b, we must draw a free body diagram of us sitting on a chair. (See attached picture.)

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N=mg-ma_{c}

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c)

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