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Ahat [919]
3 years ago
15

2.27 A gas is compressed from V1= 0.3 m3, p1=1 bar to V2= 0.1 m3, p2 =3 bar. The pressure and

Physics
1 answer:
Georgia [21]3 years ago
5 0

Answer:

-40 kJ

80 kJ

Explanation:

Work is equal to the area under the pressure vs volume graph.

W = ∫ᵥ₁ᵛ² P dV

2.27) Pressure and volume are linearly related.  When we graph P vs V, the area under the line is a trapezoid.  So the work is:

W = ½ (P₁ + P₂) (V₂ − V₁)

W = ½ (100 kPa + 300 kPa) (0.1 m³ − 0.3 m³)

W = -40 kJ

2.29) Pressure and volume are inversely proportional:

pV = k

The initial pressure and volume are 500 kPa and 0.1 m³.  So the constant is:

(500) (0.1) = k

k = 50

The final pressure is 100 kPa.  So the final volume is:

(100) V = 50

V = 0.5

The work is therefore:

W = ∫ᵥ₁ᵛ² P dV

W = ∫₀₁⁰⁵ (50/V) dV

W = 50 ln(V) |₀₁⁰⁵

W = 50 (ln 0.5 − ln 0.1)

W ≈ 80 kJ

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This could maybe help you with your answer.

4 0
3 years ago
A steel ball of mass 0.500 kg is fastened to a cord that is 70.0 cm long and fixed at the far end. The ball is then released whe
Liula [17]

Answer:

a) v₁fin = 3.7059 m/s   (→)

b) v₂fin = 1.0588 m/s     (→)

Explanation:

a) Given

m₁ = 0.5 Kg

L = 70 cm = 0.7 m

v₁in = 0 m/s   ⇒  Kin = 0 J

v₁fin = ?

h<em>in </em>= L = 0.7 m

h<em>fin </em>= 0 m   ⇒    U<em>fin</em> = 0 J

The speed of the ball before the collision can be obtained as follows

Einitial = Efinal

⇒ Kin + Uin = Kfin + Ufin

⇒ 0 + m*g*h<em>in</em> = 0.5*m*v₁fin² + 0

⇒ v₁fin = √(2*g*h<em>in</em>) = √(2*(9.81 m/s²)*(0.70 m))

⇒ v₁fin = 3.7059 m/s   (→)

b)  Given

m₁ = 0.5 Kg

m₂ = 3.0 Kg

v₁ = 3.7059 m/s    (→)

v₂ = 0 m/s

v₂fin = ?

The speed of the block just after the collision can be obtained using the equation

v₂fin = 2*m₁*v₁ / (m₁ + m₂)

⇒  v₂fin = (2*0.5 Kg*3.7059 m/s) / (0.5 Kg + 3.0 Kg)

⇒  v₂fin = 1.0588 m/s     (→)

7 0
3 years ago
Se necesita subir una carga de 500 kg (4900 N) a una altura de 1.5 m deslizándola sobre una rampa inclinada. ¿Qué longitud debe
marusya05 [52]

Answer:

4.22 m

Explanation:

Una rampa es una máquina que se utiliza para levantar un objeto con una fuerza menor a la que realmente necesitarías. Cuanto mayor sea la longitud de la rampa, menor será la magnitud de la fuerza necesaria para levantar el objeto.

Dado que:

altura de la rampa = 1.5 m, carga = 4900 N, fuerza aplicada = 1633.33 N.

La fórmula de la rampa se da como:

fuerza aplicada * longitud de la rampa = peso de la carga * altura de la rampa

1633.33 * longitud de la rampa = 4900 * 1.5

longitud de la rampa = 4900 * 1.5 / 1633.33

longitud de la rampa = 4.22 m

6 0
3 years ago
A sound wave has a frequency of 247 Hz and a wavelength of 1.4 m. What is the speed of the sound wave in air?
zavuch27 [327]
The speed of sound through air at room temperature is almost always 343 m/s. However, since it doesn't tell you that, use the equation wavelength=velocity/frequency. Plug in the numbers: 1.4=v/247, and v=345.8 m/s. 
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