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lbvjy [14]
3 years ago
10

An insulated rigid tank contains 4 kg of argon gas at 450 kPa and 30°C. A valve is now opened, and argon is allowed to escape un

til the pressure inside drops to 200 kPa. Assuming the argon remaining inside the tank has undergone a reversible, adiabatic process, determine the final mass in the tank. Answer: 2.46 kg

Physics
1 answer:
zheka24 [161]3 years ago
4 0

Answer: The final mass of the tank is 2.46kg

Explanation: All shown in the attachment.

Assumptions:

i. Argon is treated as an ideal gas at the specified conditions.

ii. Isentropic relation of ideal gas applies at the given conditions.

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A small car has a head-on collision with a large truck. Which of the following statements concerning the magnitude of the averag
andrey2020 [161]

Answer:

The small car and the truck experience the same average force.

Explanation:

Here we need to remember two of Newton's laws.

The second one says that:

F = m*a

force equals mass times acceleration.

And the third one says that;

"If an object A exerts a force on object B, then object B must exert a force of equal magnitude and opposite direction back on object A"

From the third law, if the car experiences a force F due to the impact with the truck, then the truck experiences the same force F due to the impact.

But this seems odd, because we would expect to see the car being more affected by the impact, right?

Well, this is explained by the second law.

Suppose that the mass of the car is m, and the mass of the truck is M.

such that M > m

Then for the small car we have:

F = m*a

And for the truck:

F = M*a'

Because the force is the same for both of them, we can write:

m*a = M*a'

a = (M/m)*a'

because M > m, then M/m > 1.

This means that the acceleration that the car experiences is larger than the acceleration for the truck, and this is why we would see that the car seems more affected by the impact, regardless of the fact that both vehicles experience the same force in the impact.

6 0
3 years ago
En el diseño de una sesión de aprendizaje, la profesora Patricia, tiene como propósito de aprendizaje: Aplica sus habilidades mo
Leviafan [203]

Answer:

im in hardvard dont cheat get amzing grades if you want to go here

Explanation:

5 0
3 years ago
Un engrane que gira con una velocidad de 20 rad/s, es acelerado durante 5 segundos hasta alcanzar una velocidad de 35 rad/s
larisa [96]

Answer:

a) La aceleración angular es: \alpha=2\: rad/s^{2}

b) El engranaje gira 125 radianes.

c) El engranaje hara aproximadamente 20 revoluciones.

Explanation:

a)

La aceleración angular se define como:

\alpha=\frac{\Delta \omega}{\Delta t}

Donde:

  • Δω es la diferencia de velocidad angular (en otras palabras ω(final)-ω(inicial))
  • Δt es el tiempo en el que occure el cambio de velocidad angular

\alpha=\frac{35-25}{5}

\alpha=2\: rad/s^{2}

b)

El desplazamiento angular puede ser calculado usando la siguiente ecuación:

\theta=\theta_{i}+\omega_{i}t+\frac{1}{2}\alpha t^{2}

Aqui el angulo inicial es 0, por lo tanto.

\theta=20(5)+\frac{1}{2}(2)(5)^{2}

\theta=125\: rad

El engranaje gira 125 radianes.

c)

Lo que debemos hacer aquí es convertir radianes a revoluciones.

Recordemos que 2π rad = 1 rev

Entonces:

\theta=125\: rad \times \frac{1\: rev}{2\pi\: rad}=19.89\: rev

Por lo tanto el engranaje hara aproximadamente 20 revoluciones.

Espero te haya sido de ayuda!

6 0
3 years ago
Khaled said the North star is special because it appears all over the world and its changes position.
saw5 [17]
False the North Star never changes it position
3 0
3 years ago
supose at 20 degree celsius the resistance of Tungsten thermometer is 154.9. WHen placed in a particular solution , the resistan
saw5 [17]

Answer:

T₂ = 95.56°C

Explanation:

The final resistance of a material after being heated is given by the relation:

R' = R(1 + αΔT)

where,

R' = Final Resistance = 207.4 Ω

R = Initial Resistance = 154.9 Ω

α = Temperature Coefficient of Resistance of Tungsten = 0.0045 °C⁻¹

ΔT = Change in Temperature = ?

Therefore,

207.4 Ω = 154.9 Ω[1 + (0.0045°C⁻¹)ΔT]

207.4 Ω/154.9 Ω = 1 + (0.0045°C⁻¹)ΔT

1.34 - 1 = (0.0045°C⁻¹)ΔT

ΔT = 0.34/0.0045°C⁻¹

ΔT = 75.56°C

but,

ΔT = Final Temperature - Initial Temperature

ΔT = T₂ - T₁ = T₂ - 20°C

T₂ - 20°C = 75.56°C

T₂ = 75.56°C + 20°C

<u>T₂ = 95.56°C</u>

7 0
3 years ago
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