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3 years ago

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A gas expands from an initial volume of 0.040 m^3 and an initial pressure of 210 kPa to a final volume of 0.065 m^3 while its te

mperature is kept constant. How much work is done by the system?

1 answer:

USPshnik [31]3 years ago

7 0

Answer:

286

Explanation:

p1v1/T1=p2v2/T2

then subtitude your values

T2=760*0.65*273/210*0.040

T2=135/8.4=16+273=289

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a typical cmall flashlight contains two batteries each having na emf of 2.0 v connected in series with a bulb havin ga resistanc

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Answer:

P = 0.25 W

Explanation:

Given that,

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Carbon monoxide pure substance or mixture and why

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Answer:

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2 years ago

Someone please help me answer these

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3 years ago

The main water line enters a house on the first floor. The line has a gauge pressure of 2.10 x 105 Pa. (a) A faucet on the secon

Eva8 [605]

To answer this question is necessary to apply the concepts related to Bernoulli's equation. The Bernoulli-related concept describes the behavior of a liquid moving along a streamline. Pressure can be defined as the proportional ratio between height, density and gravity:

$P = h\rho g$

Where,

h = Height

$\rho$ = Density

g = Gravity

Our values are

$\rho = 1000kg/m^3 \rightarow$ density of water at normal conditions

h = 7.3m

$g = 9.8m/s^2$

PART A) Replacing these values to find the total pressure difference we have to

$P_1 = h_1 \rho g$

$P_1 = (7.3)(1000)(9.8)$

$P_1 = 71540Pa$

In this way the pressure change would be subject to

$\Delta = P_2-P_1$

$\Delta = 2.1*10^5Pa- 0.7154*10^5Pa$

$\Delta = 138460Pa$

$\Delta = 0.135Mpa$

PART B) Considering the pressure gauge of the group as the ideal so that at a height H the water cannot flow even if it is open we have to,

$P_2 = H\rho g$

$2.1*10^5 = H (1000)(9.8)$

$H = 21.42m$

Therefore the high which could a faucet be before no water would flow from it is 21.42m

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3 years ago