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Water flows through a converging pipe at a mass flow rate of 25 kg/s. If the inside diameter of the pipes sections are 7.0 cm an

ser-zykov [4K]

Answer:

volumetric flow rate = $0.0251 m^3/s$

Velocity in pipe section 1 = $6.513m/s$

velocity in pipe section 2 = 12.79 m/s

Explanation:

We can obtain the volume flow rate from the mass flow rate by utilizing the fact that the fluid has the same density when measuring the mass flow rate and the volumetric flow rates.

The density of water is = 997 kg/m³

density = mass/ volume

since we are given the mass, therefore, the volume will be mass/density

25/997 = $0.0251 m^3/s$

volumetric flow rate = $0.0251 m^3/s$

Average velocity calculations:

<em>Pipe section A:</em>

cross-sectional area =

$\pi \times d^2\\=\pi \times 0.07^2 = 3.85\times10^{-3}m^2$

mass flow rate = density X cross-sectional area X velocity

velocity = mass flow rate /(density X cross-sectional area)

$velocity = 25/(997 \times 3.85\times10^{-3}) = 6.513m/s$

<em>Pipe section B:</em>

cross-sectional area =

$\pi \times d^2\\=\pi \times 0.05^2= 1.96\times10^{-3}m^2$

mass flow rate = density X cross-sectional area X velocity

velocity = mass flow rate /(density X cross-sectional area)

$velocity = 25/(997 \times 1.96\times10^{-3}) = 12.79m/s$

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3 years ago

.provide feedback to the controller about the robot's environment, a limited sense of sight and sound

anyanavicka [17]

Answer:

go to settings and u may figure out

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4 years ago

Read 2 more answers

Convert the improper fraction to a mixed number

Yuliya22 [10]

Answer:

18. 24/8 = 3

19. 50/12 = 4 and one-sixth

20. 18/16 = 1 and one-eighth

Explanation: I’m good at math

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3 years ago

Read 2 more answers

Consider steady one-dimensional heat conduction through a plane wall, a cylindrical shell, and a spherical shell of uniform thic

DIA [1.3K]

Answer:

Plane shell which is option b

Explanation:

The temperature in the case of a steady one-dimensional heat conduction through a plane wall is always a linear function of the thickness. for steady state dT/dt = 0

in such case, the temperature gradient dT/dx, the thermal conductivity are all linear function of x.

For a plane wall, the inner temperature is always less than the outside temperature.

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4 years ago

A motor vehicle has a mass of 1.8 tonnes and its wheelbase is 3 m. The centre of gravity of the vehicle is situated in the centr

seropon [69]

Answer:

1) The normal reactions at the front wheel is 9909.375 N

The normal reactions at the rear wheel is 8090.625 N

2) The least coefficient of friction required between the tyres and the road is 0.625

Explanation:

1) The parameters given are as follows;

Speed, u = 90 km/h = 25 m/s

Distance, s it takes to come to rest = 50 m

Mass, m = 1.8 tonnes = 1,800 kg

From the equation of motion, we have;

v² - u² = 2·a·s

Where:

v = Final velocity = 0 m/s

a = acceleration

∴ 0² - 25² = 2 × a × 50

a = -6.25 m/s²

Force, F = mass, m × a = 1,800 × (-6.25) = -11,250 N

The coefficient of friction, μ, is given as follows;

$\mu =\dfrac{u^2}{2 \times g \times s} = \dfrac{25^2}{2 \times 10 \times 50} = 0.625$

Weight transfer is given as follows;

$W_{t}=\dfrac{0.625 \times 0.9}{3}\times \dfrac{6.25}{10}\times 18000 = 2109.375 \, N$

Therefore, we have for the car at rest;

Taking moment about the Center of Gravity CG;

$F_R$ × 1.3 = 1.7 × $F_F$

$F_R$ + $F_F$ = 18000

$F_R + \dfrac{1.3 }{1.7} \times F_R = 18000$

$F_R$ = 18000*17/30 = 10200 N

$F_F$ = 18000 N - 10200 N = 7800 N

Hence with the weight transfer, we have;

The normal reactions at the rear wheel $F_R$ = 10200 N - 2109.375 N = 8090.625 N

The normal reactions at the front wheel $F_F$ = 7800 N + 2109.375 N = 9909.375 N

2) The least coefficient of friction, μ, is given as follows;

$\mu = \dfrac{F}{R} = \dfrac{11250}{18000} = 0.625$

The least coefficient of friction, μ = 0.625.

3 0

4 years ago