Un vehículo parte del reposo y acelera constantemente a 6 m/s2

Carbon dioxide in a piston-‐‐cylinder is expanded in a polytropic manner. The initialtemperature and pressure are 400 K and 550

Arisa [49]

Answer:

q_poly = 14.55 KJ/kg

Explanation:

Given:

Initial State:

P_i = 550 KPa

T_i = 400 K

Final State:

T_f = 350 K

Constants:

R = 0.189 KJ/kgK

k = 1.289 = c_p / c_v

n = 1.2 (poly-tropic index)

Find:

Determine the heat transfer per kg in the process.

Solution:

-The heat transfer per kg of poly-tropic process is given by the expression:

q_poly = w_poly*(k - n)/(k-1)

- Evaluate w_poly:

w_poly = R*(T_f - T_i)/(1-n)

w_poly = 0.189*(350 - 400)/(1-1.2)

w_poly = 47.25 KJ/kg

-Hence,

q_poly = 47.25*(1.289 - 1.2)/(1.289-1)

q_poly = 14.55 KJ/kg

4 0

3 years ago

If 1.00 mol of argon is placed in a 0.500-L container at 28.0 ∘C , what is the difference between the ideal pressure (as predict

Rudik [331]

Answer:

1.98 atm

Explanation:

Given that:

Temperature = 28.0 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T₁ = (28 + 273.15) K = 301.15 K

n = 1

V = 0.500 L

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L atm/ K mol

Applying the equation as:

P × 0.500 L = 1 ×0.0821 L atm/ K mol × 301.15 K

⇒P (ideal) = 49.45 atm

Using Van der Waal's equation

$\left(P+\frac{an^2}{V^2}\right)\left(V-nb\right)=nRT$

R = 0.0821 L atm/ K mol

Where, a and b are constants.

For Ar, given that:

So, a = 1.345 atm L² / mol²

b = 0.03219 L / mol

So,

$\left(P+\frac{1.345\times \:1^2}{0.500^2}\right)\left(0.500-1\times 0.03219\right)=1\times 0.0821\times 301.15$

$P+\frac{1.345}{0.25}=\frac{24.724415}{0.46781}$

$P=\frac{24.724415}{0.46781}-\frac{1.345}{0.25}$

⇒P (real) = 47.47 atm

Difference in pressure = 49.45 atm - 47.47 atm = 1.98 atm

4 0

2 years ago

These graphs show measurements of the density of air as different sound waves pass a single point.

Ymorist [56]

Answer: D!

It is the option with the greatest amplitude.

5 0

3 years ago

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A laser source has a bandwidth of 30GHz (a) Calculate the coherence length of the source b) What is the time separation of secti

abruzzese [7]

Explanation:

It is given that,

Bandwidth of a laser source, $f=30\ GHz=30\times 10^9\ Hz$

(b) Let t is the time separation of sections of sections of the light wave that can still interfere. The time period is given by :

$T=\dfrac{1}{f}$

$T=\dfrac{1}{30\times 10^9}$

$T=3.33\times 10^{-11}\ s$

(a) Let h is the coherence length of the source. It is given by :

$l=c\times T$

c is the speed of light

$l=3\times 10^8\times 3.33\times 10^{-11}$

l = 0.0099 m

Hence, this is the required solution.

6 0

2 years ago

Which equation is correct according to Ohm’s law?

Korvikt [17]

You didn't give us anything other than the question. No options are provided so I cannot answer. Nobody can.

4 0

2 years ago

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