3 years ago

How does the start function make your programs easier to understand?

Engineering

1 answer:

ad-work [718]3 years ago

7 0

Explanation:

All of the commands you wish to execute must go inside of the start function, between the curly braces.

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What types of issues MAY occur to slow or prevent the best outcome?

german

Answer:

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3 years ago

Can i have answer of this question please?

cestrela7 [59]

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2 years ago

The Reynolds number is a dimensionless quantity used in fluid mechanics. It is defined as:where d is the pipe diameter, v is the

ELEN [110]

Lb(force)•second/ft^2

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3 years ago

Steam enters a turbine operating at steady state with a mass flow of 10 kg/min, a specific enthalpy of 3100 kJ/kg, and a velocit

Natali [406]

Answer:

$\dot W_{out} = 133.327\,kW$

Explanation:

The model for the turbine can be derived by means of the First Law of Thermodynamics:

$-\dot Q_{out}-\dot W_{out} +\dot m \cdot \left[(h_{in}-h_{out})+\frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2}) + g\cdot (z_{in}-z_{out})\right] =0$

The work produced by the turbine is:

$\dot W_{out}=-\dot Q_{out} +\dot m \cdot \left[(h_{in}-h_{out})+\frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2}) + g\cdot (z_{in}-z_{out})\right]$

The mass flow and heat transfer rates are, respectively:

$\dot m = (10\frac{kg}{min})\cdot (\frac{1\,min}{60\,s} )$

$\dot m = 0.167\,\frac{kg}{s}$

$\dot Q_{out} = (0.167\,\frac{kg}{s} )\cdot (1.1\times 10^{3}\,\frac{J}{kg} )$

$\dot Q_{out} = 183.7\,W$

Finally:

$\dot W_{out} = -183.7\,W + (0.167\,\frac{kg}{s} )\cdot \left(8\times 10^{5}\,\frac{J}{kg} -562,5\,\frac{J}{kg} +29.43\,\frac{J}{kg} \right)$

$\dot W_{out} = 133.327\,kW$

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3 years ago

Viteza unui mobil care se deplasează cu accelerație constantă crește de la 3,2 m/s la 5,2 m/s în timp de 8 s. Accelerația mobilu

Verizon [17]

Answer:

$a=0.25\ m/s^2$

Explanation:

Initial speed of the mobile = 3.2 m/s

Final speed of the mobile = 5.2 m/s

Time, t = 8 s

We need to find the acceleration of the mobile. It can be given by the change in velocity divided by time. So,

$a=\dfrac{v-u}{t}\\\\a=\dfrac{(5.2-3.2)\ m/s}{8\ s}\\\\=0.25\ m/s^2$

So, the acceleration of the mobile is $0.25\ m/s^2$ .

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2 years ago