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IrinaK [193]
3 years ago
11

PS Final Exam

Physics
1 answer:
ivolga24 [154]3 years ago
4 0
A rolling friction jdisns
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Cuanto cambia la entropía de 0.50 kg de vapor de mercurio [Lv: 2.7 x 10⁵ j/kg ] al calentarse en su punto de ebullición de 357°
lord [1]

Answer:

La entropía del vapor de mercurio cambia en 214.235 joules por Kelvin.

Explanation:

Por definición de entropía (S), medida en joules por Kelvin, tenemos la siguiente expresión:

dS = \frac{\delta Q}{T} (1)

Donde:

Q - Ganancia de calor, en joules.

T - Temperatura del sistema, en Kelvin.

Ampliamos (1) por la definición de calor latente:

dS = \frac{L_{v}}{T}\cdot dm (1b)

Donde:

m - Masa del sistema, en kilogramos.

L_{v} - Calor latente de vaporización, en joules

Puesto que no existe cambio en la temperatura durante el proceso de vaporización, transformamos la expresión diferencial en expresión de diferencia, es decir:

\Delta S = \frac{\Delta m \cdot L_{v}}{T}

Como vemos, el cambio de la entropía asociada al cambio de fase del mercurio es directamente proporcional a la masa del sistema. Si tenemos que m = 0.50\,kg,L_{v} = 2.7\times 10^{5}\,\frac{J}{kg} and T = 630.15\,K, entonces el cambio de entropía es:

\Delta S = \frac{(0.50\,kg)\cdot \left(2.7\times 10^{5}\,\frac{J}{kg} \right)}{630.15\,K}

\Delta S = 214.235 \,\frac{J}{K}

La entropía del vapor de mercurio cambia en 214.235 joules por Kelvin.

3 0
3 years ago
What is the stretch when you pull with a force of 25 N on a spring with a spring constant of 8 N/m? *
Pani-rosa [81]

Hooke's Law

\tt F=k.\Delta x

k = spring constant

x = stretch

F = force

Input the value

\tt \Delta x=\dfrac{F}{k}=\dfrac{25}{8}=3.125\rightarrow 3.13\:m

7 0
2 years ago
The total resistance of resistors in parallel is always _________ the sum of the resistors.
docker41 [41]
A. less than

In fact, the resistance will be less than that of the lowest resistance resistor.
7 0
3 years ago
Read 2 more answers
hat are the wavelengths of peak intensity and the corresponding spectral regions for radiating objects at (a) normal human body
bagirrra123 [75]

Answer:

(a)

\lambda _{m}=9.332 \times 10^{-6}m

(b)

\lambda _{m}=1.632 \times 10^{-6}m

(c) \lambda _{m}=4.988 \times 10^{-7}m

 

Explanation:

According to the Wein's displacement law

\lambda _{m}\times T = b

Where, T be the absolute temperature and b is the Wein's displacement constant.

b = 2.898 x 10^-3 m-K

(a) T = 37°C = 37 + 273 = 310 K

\lambda _{m}=\frac{b}{T}

\lambda _{m}=\frac{2.893\times 10^{-3}}{310}

\lambda _{m}=9.332 \times 10^{-6}m

(b) T = 1500°C = 1500 + 273 = 1773 K

\lambda _{m}=\frac{b}{T}

\lambda _{m}=\frac{2.893\times 10^{-3}}{1773}

\lambda _{m}=1.632 \times 10^{-6}m

(c) T = 5800 K

\lambda _{m}=\frac{b}{T}

\lambda _{m}=\frac{2.893\times 10^{-3}}{5800}

\lambda _{m}=4.988 \times 10^{-7}m

5 0
3 years ago
An object in space is initially stationary relative to the Earth. Then, a force begins acting on the object, starting with a for
11111nata11111 [884]

Answer:

v_f =63 m/s

Explanation:

given,

starting force = 0 N

uniform rate increase to 36 N

time of action of Force = 35 s

mass of the body = 10 Kg

Speed of the object = ?

From the given data

if we plot F-t curve  we will get a triangular shape

we know,

Impulse = Area between F-t curve

            = (1/2) x base x height

             = 0.5 x 35 x 36

             = 630 N.s

now use Impulse-momentum theorem

Impulse = change in momentum

630 = 10 x (v_f - vi)

630 = 10 x (v_f - 0)

v_f =63 m/s

Speed of the object at 35 sec is equal to v_f =63 m/s

8 0
3 years ago
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