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nordsb [41]
3 years ago
9

In the pacific northwest orographic precipitation usually occurs where?

Physics
1 answer:
VMariaS [17]3 years ago
4 0
That’s the answer hope this helps!!!!

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how much water is needed to produce 1kwh of electricity at a power plant that is 30% efficient if the temperature increase 10 C
Dimas [21]

The amount of water needed is 287 kg

Explanation:

The amount of energy that we need to produce with the power plant is

E=1 kWh = (1000W)(1h)=(1000W)(3600s)=3.6\cdot 10^6 J

We also know that the power plant is only 30% efficient, so the energy produced in input must be:

E_{in}=\frac{E}{0.30}=\frac{3.6\cdot 10^6}{0.3}=1.2\cdot 10^7 J

The amount of water that is needed to produce this energy can be found using the equation

E_{in}=mC\Delta T

where:

m is the amount of water

C=4186 J/kg^{\circ}C is the specific heat capacity of water

\Delta T=10^{\circ}C is the increase in temperature

And solving for m, we find:

m=\frac{E_{in}}{C\Delta T}=\frac{1.2\cdot 10^7}{(4186)(10)}=287 kg

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

3 0
2 years ago
________________ can carry thermal energy from object to object, even through empty space.
sdas [7]
Radiation can carry thermal energy.
8 0
3 years ago
Una carga positiva de 4 x 10-5 C, se encuentra a 0.05 m de otra carga positiva de 2 x 10-5 C. Calcular la fuerza que se ejerce e
KATRIN_1 [288]

Answer:

La fuerza que se ejerce entre las dos cargas es 2880 N.

Explanation:

La ley de Coulomb indica que los cuerpos cargados sufren una fuerza atractiva o repulsiva al acercarse. La fuerza es atractiva si las cargas son del signo opuesto y repulsión si son del mismo signo. El valor de la fuerza es proporcional al producto del valor de sus cargas e inversamente proporcional al cuadrado de la distancia que los separa. Esto se expresa matemáticamente como:

F=k*\frac{Q*q}{r^{2} }

donde:

  • F es la fuerza eléctrica de atracción o repulsión. Se mide en Newtons (N).
  • Q y q son los valores de las dos cargas puntuales. Se miden en culombios (C).
  • r es el valor de la distancia que los separa. Se mide en metros (m).
  • k es una constante de proporcionalidad llamada constante de la ley de Coulomb.

En este caso:

  • F= ?
  • Q= 4*10⁻⁵ C
  • q= 2*10⁻⁵ C
  • r= 0.05 m
  • k= 9*10⁹ \frac{N*m^{2} }{C^{2} }

Reemplazando:

F=9*10^{9} \frac{N*m^{2} }{C^{2} }*\frac{4*10^{-5} C*2*10^{-5}C }{(0.05 m)^{2} }

F= 2880 N

<u><em>La fuerza que se ejerce entre las dos cargas es 2880 N.</em></u>

7 0
3 years ago
what is the final volume of a gas with an initial volume of 200 .0ml if the pressure decreases from 500.0 kpa to 250.0 kpa
Serjik [45]

<u>Answer:</u>

Given Data:

V2 ?

V1 = 200 ml,

P1 = 500kpa and P2 = 2500kpa


From <em>Boyels law</em> (Ideal gas law) Where temperature remains constant, <em>Pressure is inversely proportional to volume.</em>

                   P1 .V1 = P2 . V2  ;       T=Constant

                        V2 = (P1.V1) ÷ P2

                              = (500×200) ÷(250)

                              = 400 ml

<em>The final volume of the gas is 400 ml.</em>

<em>Note: Increased pressure decreases the volume or decreased pressure increase the volume as they are inversely proportional.</em><em> Here in our answer pressure decreases from 500 kpa to 250 kpa so volume increases from 200 ml to 400 ml. </em>

5 0
2 years ago
Usually wire resistance can be neglected. But we see the effect of it sometimes when a light dims as another high-current device
tangare [24]

Please find the figure attached below:

Answer:

a.Resistance of bulb=R_{2}=192 ohm

b.power consumed by bulb after motor disconnection=74.609 W

Explanation:

<u>a.Resistance of bulb=</u>R_{2}<u>=?</u>

As we know that P=\frac{P^{2} }{R}

putting R as resistance of bulb i.e. R=R_{2}

P=\frac{V^{2}}{R_{2} }\\R_{2}=\frac{V^{2}}{P} \\ R_{2}=\frac{(120)^{2} }{75}\\ R_{2}=192ohm

<u>b.power consumed by bulb after motor disconnection? </u>

from the figure we see that R_{1}\ and\ R_{2} are in series so

R_{eq}=R_{1} +R_{2} \\R_{eq}=192+0.5\\R_{eq}=192.5ohm

current through their resistance is

I_{eq}=\frac{V}{R_{eq} }\\ I_{eq}=\frac{120}{192.5 } \\I_{eq}=0.6233A

Power consumed by bulb is

P_{b} =I^{2}R\\\\ P_{b} =(0.6233)^{2}(192)\\\\P_{b} =74.609W

3 0
3 years ago
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