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Nonamiya [84]
3 years ago
5

In Kepler's model of the solar system, the planets move around the Sun in _________ orbits.

Physics
1 answer:
cestrela7 [59]3 years ago
3 0
B. elliptical.

The planets move in elliptical orbits round the Sun.
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Where are chromosomes found in an animal cell?
timama [110]
Chromosomes<span> are thread-like structures </span>located<span> inside the nucleus of </span>animal<span> and plant </span>cells<span>.</span>
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A friend rides, in turn, the rims of three fast merry-go-rounds while holding a sound source that emits isotropically at a certa
Aliun [14]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

The Ranking of the curve according to their speed would be equal Rank because    v_1 =v_2 =v_3

b

 The first frequency would have a higher rank compared to the other two which will have the same ranking when ranked with respect to their angular velocities because

                                w_1 >w_2 = w_3  

c

The ranking of  the second third frequency would be the same but their ranking would be greater than that of the first frequency because

                          r_2 =r_3 >r_1

Explanation:

Mathematically Frequency can be represented as

                         F = \frac{v}{\lambda}

Where \lambda is the wavelength and v is the velocity

   Now looking at the diagram we see that

          For the  first frequency we have

             Let the wavelength be  \lambda_1 = \lambda , and the frequency  F_1 = F

           For  the second frequency

           Let the wavelength be  \lambda_2 = 2 \lambda , and the frequency F_2 = \frac{F}{2}

           For  the third frequency

           Let the wavelength be  \lambda_3 = 2\lambda ,  and the frequency F_3 = \frac{F}{2}

To obtain v for each of the frequency we make v the subject in the equation above for each frequency

  So,

        For the  first frequency we have

                                 v_1 = \lambda_1 F_1 = \lambda F

          For  the second frequency

                               v_2 = \lambda_2 F_2 = 2 \lambda*\frac{F} {2} = \lambda F      

           For  the third frequency

                               v_3 = \lambda_3 F_3 = 2 \lambda*\frac{F} {2} = \lambda F

Hence

The Ranking of the curve according to their speed would be equal Rank because    v_1 =v_2 =v_3

 Mathematically angular speed can be represented as

                           w = 2 \pi f

   For the  first frequency we have

                          w_1 = 2\pi F_1 = 2 \pi F                        

    For  the second frequency

                        w_2 = 2 \pi F_2 = 2 \pi \frac{F}{2}  = \pi F

     For  the third frequency

                      w_3 = 2 \pi F_3 = 2 \pi \frac{F}{2}  = \pi F  

 Hence

          The first frequency would have a higher rank compared to the other two which will have the same ranking when ranked with respect to their angular velocities because

                                w_1 >w_2 = w_3  

Mathematically the relationship between the angular velocity and the linear velocity can be represented as

                            v = wr

                    =>    r = \frac{v}{w}

 Since the linear velocity is constant we have that

                            r \  \alpha \  \frac{1}{w}

This means that r varies inversely to the angular velocity ,What this means for ranking due to the radius is that the ranking of  the second third frequency would be the same but their ranking would be greater than that of the first frequency because

                          r_2 =r_3 >r_1

       

5 0
3 years ago
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
ElenaW [278]

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

Fluid density, \rho_{f} = 990\ kg/m^{3}

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.

Now,

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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