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storchak [24]
3 years ago
11

4. Compara. La composición de la gasolina para los coches cambia del invierno al verano. La mezcla de los componentes de la gaso

lina debe evitar problemas en su vaporización en los cilindros del motor. Por eso, las compañías petrolíferas preparan distintas mezclas para el invierno o para el verano: una de las mezclas tiene más componentes volátiles que la otra. Argumenta cuál tendrá más componentes volátiles en su composición y qué problemas surgirían si se emplearan al revés, la de invierno en verano y la de verano en invierno.
Physics
1 answer:
dedylja [7]3 years ago
5 0
Skisosjssnbxhxsndjsksksksa
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Why do you not come to thermal equilibrium on a cold day
Tju [1.3M]

Two physical systems are in thermal equilibrium if no heat flows between them when they are connected by a path permeable to heat. Thermal equilibrium obeys the zeroth law of thermodynamics. A system is said to be in thermal equilibrium with itself if the temperature within the system is spatially and temporally uniform.

Systems in thermodynamic equilibrium are always in thermal equilibrium, but the converse is not always true. If the connection between the systems allows transfer of energy as heat but does not allow transfer of matter or transfer of energy as work, the two systems may reach thermal equilibrium without reaching thermodynamic equilibrium.

3 0
4 years ago
A solenoid has a radius Rs = 14.0 cm, length L = 3.50 m, and Ns = 6500 turns. The current in the solenoid decreases at the rate
babymother [125]

Answer:

E = 58.7 V/m

Explanation:

As we know that flux linked with the coil is given as

\phi = NBA

here we have

A = \pi R_s^2

B = \mu_o N i/L

now we have

\phi = N(\mu_o N i/L)(\pi R_s^2)

now the induced EMF is rate of change in magnetic flux

EMF = \frac{d\phi}{dt} = \mu_o N^2 \pi R_s^2 \frac{di}{dt}/L

now for induced electric field in the coil is linked with the EMF as

\int E. dL = EMF

E(2\pi r_c) = \mu_o N^2 \pi R_s^2 \frac{di}{dt}/L

E = \frac{\mu_o N^2 R_s^2 \frac{di}{dt}}{2 r_c L}

E = \frac{(4\pi \times 10^{-7})(6500^2)(0.14^2)(79)}{2(0.20)(3.50)}

E = 58.7 V/m

3 0
3 years ago
You connected the 5 Ω, 10 Ω, 15 Ω resistors in series with a 90 V battery. What is the current?​
kykrilka [37]

Answer:

3A

Explanation:

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I=V/R 90/30

I=3

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The dog’s speed is
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I need help on the data section of the circuit design lab on Edg.
Arte-miy333 [17]

I hope it's not too late, but here you go

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