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Verizon [17]
3 years ago
10

Un neumático sin cámara, soporta una presión de 1.5 atm cuando la temperatura ambiente es de 300°K. ¿Qué presión llegará a sopor

tar dicho neumático si, en el transcurso de un viaje, las ruedas alcanzan una temperatura de 67ºC? Atm
Physics
1 answer:
arlik [135]3 years ago
4 0

Answer:

El neumático soportará una presión de 1.7 atm.

Explanation:

Podemos encontrar la presión final del neumático usando la ecuación del gas ideal:

PV = nRT

En donde:

P: es la presión

V: es el volumen

n: es el número de moles del gas

R: es la constante de gases ideales

T: es la temperatura

Cuando el neumático soporta la presión inicial tenemos:

P₁ = 1.5 atm

T₁ = 300 K

V_{1} = \frac{nRT_{1}}{P_{1}}  (1)  

La presión cuando T = 67 °C es:

P_{2} = \frac{nRT_{2}}{V_{2}}   (2)

Dado que V₁ = V₂  (el volumen del neumático no cambia), al introducir la ecuación (1) en la ecuación (2) podemos encontrar la presión final:

P_{2} = \frac{nRT_{2}}{V_{2}} = \frac{nRT_{2}}{\frac{nRT_{1}}{P_{1}}} = \frac{P_{1}T_{2}}{T_{1}} = \frac{1.5 atm*(67 + 273)K}{300 K} = 1.7 atm  

Por lo tanto, si en el transcurso de un viaje las ruedas alcanzan una temperatura de 67 ºC, el neumático soportará una presión de 1.7 atm.

Espero que te sea de utilidad!

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A child throws a ball with an initial speed of 8.00 m/s at an angle of 40.0° above the horizontal. The ball leaves her hand 1.00
vivado [14]

Answer:

7.5 m

Explanation:

v = initial speed of the ball = 8 m/s

\theta = angle of launch = 40° deg

Consider the motion along the vertical direction :

v_{oy} = initial velocity along vertical direction = v Sin\theta = 8 Sin40 = 5.14 m/s

a_{y} = acceleration along vertical direction = - 9.8 m/s²

t = time of travel

y = vertical displacement = - 1 m

Using the kinematics equation

y = v_{oy}t + (0.5)a_{y}t^{2}

- 1 = (5.14)t + (0.5)(- 9.8)t^{2

t = 1.22 sec

Consider the motion along the horizontal direction :

v_{ox} = initial velocity along horizontal direction = v Sin\theta = 8 Cos40 = 6.13 m/s

a_{x} = acceleration along vertical direction = 0 m/s²

t = time of travel = 1.22 sec

x = horizontal displacement = ?

Using the kinematics equation

x = v_{ox}t + (0.5)a_{x}t^{2}

x = (6.13)(1.22) + (0.5)(0)(1.22)^{2

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4 0
3 years ago
4.39 moles of gas in a box has a pressure of 2.25 atm at temperature of 385K. What is the volume of the box?
bazaltina [42]

Answer:

0.0619 m^3

Explanation

number of moles = n = 4.39 mol

pressure = P = 2.25 atm =2.25×1.01×10^5 Pa= 2.27×10^5 Pa

Molar gas constant =R = 8.31 J/(mol K)

Temperature T= 385K

volume of gas = V =?

BY GENERAL GAS LAW WE HAVE

PV = nRT

or V = nRT/P

or V = (4.39×8.31×385)/(2.27×10^5)

V = 0.0618728

V =  0.0619 m^3

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1) 3 miles/Hour

The speed is defined as the distance covered divided by the time taken:

v=\frac{d}{t}

where

d = 1.5 mi is the distance

t = 0.5 h is the time taken

Substituting,

v=\frac{1.5}{0.5}=3 mi/h

2) 1.34 m/s south

Velocity, instead, is a vector, so it has both a magnitude and a direction. We have:

d=1.5 mi \cdot 1609 m/mi = 2414 m is the displacement in meters

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Substituting,

v=\frac{2414 m}{1800 s}=1.34 m/s

And the direction of the velocity is the same as the displacement, so it is south.

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