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nasty-shy [4]
3 years ago
5

In order to increase the amount of work completed it is necessary to

Physics
2 answers:
Dimas [21]3 years ago
5 0
Increase the input energy to the system
navik [9.2K]3 years ago
3 0
Do all you can in one big day that you have time off or work on one thing then work on the other at the same time

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A circular grill of diameter 0.25 m has an emissivity of 0.8. If the surface temperature is maintained at 150°C, determine the r
Schach [20]

Answer:

the required electrical power when the room air and surroundings are at 30°C.= 52.51822 Watt

Explanation:

Power required to maintain the surface temperature at 150°C from 20°C

P= εσA(T^4-t^4)

P= power in watt

ε= emissivity

A=  area of surface

T= 150°C= 423 K

t= 20°C= 303K

/sigma= 5.67×10^{-8} watt/m^2K^4

putting vales we get

= 0.8\times5.67\times10^{-8} \pi\frac{0.25^4}{4}(423^4-303^4)

P=52.51822 Watt

the required electrical power when the room air and surroundings are at 30°C.= 52.51822 Watt

4 0
2 years ago
NEED ASAP PAST DUE
iVinArrow [24]

Answer: The terrestrial planets, Mars, Earth, Venus, and Mercury all have relatively high densities and low gas content, e.g., they are small and rocky. The Jovian (or giant planets), Jupiter, Saturn, Uranus, and Neptune, are very large and have rather low densities, e.g., they are gaseous.

Explanation:

:)

4 0
3 years ago
Air enters an adiabatic compressor at 104 kPa and 292 K and exits at a temperature of 565 K. Determine the power (kW) for the co
ladessa [460]

Answer:

\dot W_{in} = 49.386\,kW

Explanation:

An adiabatic compressor is modelled as follows by using the First Law of Thermodynamics:

\dot W_{in} + \dot m \cdot c_{p}\cdot (T_{1}-T_{2}) = 0

The power consumed by the compressor can be calculated by the following expression:

\dot W_{in} = \dot m \cdot c_{v}\cdot (T_{2}-T_{1})

Let consider that air behaves ideally. The density of air at inlet is:

P\cdot V = n\cdot R_{u}\cdot T

P\cdot V = \frac{m}{M}\cdot R_{u}\cdot T

\rho = \frac{P\cdot M}{R_{u}\cdot T}

\rho = \frac{(104\,kPa)\cdot (28.02\,\frac{kg}{kmol})}{(8.315\,\frac{kPa\cdot m^{3}}{kmol\cdot K} )\cdot (292\,K)}

\rho = 1.2\,\frac{kg}{m^{3}}

The mass flow through compressor is:

\dot m = \rho \cdot \dot V

\dot m = (1.2\,\frac{kg}{m^{3}})\cdot (0.15\,\frac{m^{3}}{s} )

\dot m = 0.18\,\frac{kg}{s}

The work input is:

\dot W_{in} = (0.18\,\frac{kg}{s} )\cdot (1.005\,\frac{kJ}{kg\cdot K})\cdot (565\,K-292\,K)

\dot W_{in} = 49.386\,kW

5 0
3 years ago
A 70.-kilogram cyclist develops 210 watts of
Temka [501]
     The Definition of Potence is given by:

P=Fv
 
     Entering the unknowns:

P=Fv \\ 210=7F \\ F= \frac{210}{7}  \\ \boxed {F=30N}

Number 2

If you notice any mistake in my english, please let me know, because i am not native.

8 0
3 years ago
!Necesito ayuda:Un movil parte del reposo y en 5seg adquiere una rapidez de 12m/seg.Calcular la distancia recorrida.
KatRina [158]

Answer:

Distancia = 60 metros

Explanation:

Dados los siguientes datos;

Velocidad inicial = 0 m/s (ya que comienza desde el reposo)

Velocidad final = 12 m/s

Por lo tanto, velocidad total = velocidad inicial + velocidad final

Velocidad total = 0 + 12 = 12 m/s

Tiempo = 5 segundos

Para encontrar la distancia recorrida;

Distancia = velocidad total * tiempo

Sustituyendo en la fórmula, tenemos;

Distancia = 12 * 5

Distancia = 60 metros

7 0
3 years ago
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