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Natasha2012 [34]
3 years ago
13

A scooter is accelerated from rest at the rate of 8m/s . How long will it take to cover a distance of 32m?​

Physics
1 answer:
TiliK225 [7]3 years ago
6 0
  • <em>S</em><em>peed </em><em>=</em><em> </em><em>8</em><em> </em><em>m/</em><em>s</em>
  • <em>Distance</em><em> </em><em>=</em><em> </em><em>3</em><em>2</em><em> </em><em>m</em>

<u>We need to find time</u>

<h3>We know that ,</h3>

\begin{gathered} {\underline{\boxed{ \rm {\red{Speed \:  =  \:  \frac{Distance}{Time} }}}}}\end{gathered}

<h3>So ,</h3>

\begin{gathered} {\underline{\boxed{ \rm {\orange{Time \:  =  \:  \frac{Distance}{Speed} }}}}}\end{gathered}

<h3><u>Substuting</u><u> the</u><u> values</u></h3>

\begin{cases} \large\rm{\purple{ {\leadsto} }} \:  \: Time \:  =  \:  \frac{32}{8} \\  \\ \large\rm{\purple{ {\leadsto} }} \:  \: Time \:  =  \:   \cancel\frac{32}{8} \\  \\ \large\rm{\purple{ {\leadsto} }} \:  \: Time \:  =  \:  4 \: seconds \end{cases}

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A fluid is flowing through a circulat tube at 0.4 kg/s. Tube inner surface is smooth with a diameter 0.014 m. Fluid density is 9
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Answer:

The convection coefficient is 15456.48\ W/m^{2}K

Solution:

Mass flow rate, \dot{m} = 0.4\ kg

Inner diameter of the tube, d = 0.014 m

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Specific Heat, C = 3845 J/K

Thermal Conductivity, K = 0.74

Prandtl Number, P_{r} = 8.6

Heat flux, \dot{q} = 71,297\ W/m^{2}

Viscosity, \mu = 0.00079\ Ns/m^{2}

Now,

To calculate the convection heat coefficient, h:

Determine the cross sectional area of the circular tube:

A_{c} = \frac{\pi}{4}d^{2} = \frac{\pi}{4}\times (0.014)^{2} = 1.54\time 10^{- 4}\ m^{2}

Determine the velocity of the fluid inside the tube by mass flow rate:

\dot{m} = \rho_{f}A_{c}v

0.4 = 990\times 1.54\time 10^{- 4}v

v = 2.624 m/s

Determine the Reynold's Number, R_{e}:

R_{e} = \frac{\rho_{f}dv}{\mu}

R_{e} = \frac{990\times 0.014\times 2.624}{0.00079} = 46036.253

Thus it is clear that R_{e} > 10,000 hence flow is turbulent.

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Determine the Nusselt Number:

N_{u} = 0.023R_{e}^{0.8}P_{r}^{0.4}

N_{u} = 0.023\times 46036.253^{0.8}\times 8.6^{0.4} = 292.42

Also,

N_{u} = \frac{dh}{K}

where

h = convection coefficient

Now,

292.42 = \frac{0.014\times h}{0.74}

h = 15456.48\ W/m^{2}K

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