Ask question

Ask question

All categories

3 years ago

9

An automobile tire having a temperature of 3.4 ◦C (a cold tire on a cold day) is filled to a gauge pressure of 24 lb/in2 . What

would be the gauge pressure in the tire when its temperature rises to 26◦C? For simplicity, assume that the volume of the tire remains constant, that the air does not leak out and that the atmospheric pressure remains constant at 14.7 lb/in2 . Answer in units of lb/in2 .

2 answers:

dusya [7]3 years ago

6 0

Answer:

27.16 lb/in²

Explanation:

initial temperature, T1 = 3.4 °C = 276.4 K

initial gauge pressure, P1 = 24 lb/in²

atmospheric pressure, Po = 14.7 lb/in²

initial absolute pressure, P1' = Po + P1 = 14.7 + 24 = 38.7 lb/in²

final temperature, T2 = 26 °C = 299 K

Let the final gauge pressure is P2.

use the ideal gas equation and the volume is constant.

$\frac{P_{1}'}{T_{1}}=\frac{P_{2}'}{T_{2}}$

$\frac{38.7}{276.4}=\frac{P_{2}'}{299}$

P2' = 41.86 lb/in²

Now the gauge pressure, P2 = P2' - Po = 41.86 - 14.7 = 27.16 lb/in²

Thus, the new gauge pressure is 27.16 lb/in².

notka56 [123]3 years ago

4 0

Answer:

$27.164 lb/in^2$

Explanation:

We are given that

Gauge pressure at 3.4 degree Celsius,P= $24lb/in^2$

We have to find the gauge pressure in tire when the temperature rises to 26 degree Celsius.

Atmospheric pressure= $14.7lb/in^2$

$P_1=P+14.7=24+14.7=38.7lb/in^2$

K=273+Degree Celsius

$T_1=3.4+273=276.4 K$

$T_2=26+273=299K$

$P_2=\frac{P_1T_2}{T_1}$

$P_2=\frac{38.7\times 299}{276.4}$

$P_2=41.864 lb/in^2$

Gauge pressure in tire when the temperature rises to 26 degree Celsius.= $41.864-14.7=27.164lb/in^2$

You might be interested in

Whenever a musician plays a guitar, they pluck one of the guitar strings to produce a standing wave in the string. Then pinch do

UkoKoshka [18]

Answer:

It's due to the distance from either ends of strings origin...

Explanation:

As we know that waves behave moving in a flow from one side to another side and this gives a prospective of motion. Suppose a wave is pinched from the near one end of a guitar then due to the distortion created by the point of tie of strings the wave super imposes and moves with a velocity v and produces a wave frequency f. as we the pinching go down to the center the wave stabilizes itself to a stationary origin right at the center and the frequency then changes accordingly as moving down on the string.

7 0

3 years ago

The Ha line of the Balmer series is emitted in the transition from n-3 to n 2. Compute the wavelength of this line for H and 2H.

Citrus2011 [14]

Explanation:

According to Rydberg's formula, the wavelength of the balmer series is given by:

$\frac{1}{\lambda}=R(\frac{1}{2^2}-\frac{1}{3^2})$

R is Rydberg constant for an especific hydrogen-like atom, we may calculate R for hydrogen and deuterium atoms from:

$R=\frac{R_{\infty}}{(1+\frac{m_e}{M})}$

Here, $R_{\infty}$ is the "general" Rydberg constant, $m_e$ is electron's mass and M is the mass of the atom nucleus

For hydrogen, we have, $M=1.67*10^{-27}kg$ :

$R_H=\frac{1.09737*10^7m^{-1}}{(1+\frac{9.11*10^{-31}kg}{1.67*10^{-27}kg})}\\R_H=1.09677*10^7m^{-1}$

Now, we calculate the wavelength for hydrogen:

$\frac{1}{\lambda}=R_H(\frac{1}{2^2}-\frac{1}{3^2})\\\lambda=[R_H(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=[1.0967*10^7m^{-1}(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=6.5646*10^{-7}m=656.46nm$

For deuterium, we have $M=2(1.67*10^{-27}kg)$ :

$R_D=\frac{1.09737*10^7m^{-1}}{(1+\frac{9.11*10^{-31}kg}{2*1.67*10^{-27}kg})}\\R_D=1.09707*10^7m^{-1}\\\\\lambda=[R_D(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=[1.09707*10^7m^{-1}(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=6.5629*10^{-7}=656.29nm$

5 0

3 years ago

Where did the carbon in all living things come from?

Maru [420]

D (it was produced by cyanobacteria)!!!

3 0

3 years ago

Read 2 more answers

(I) A car slows down from 28 m????s to rest in a distance of 88 m. What was its acceleration, assumed constant?

Nataly [62]

Answer:

The value is $a = - 4.45 m/s^2$

Explanation:

From the question we are told that

The initial speed is $u = 28 \ m/s$ at a distance of $s_1 = 0 \ m$

The final speed is $v = 0 \ m/s$ at a distance of $s_2 = 88 \ m$

Generally from the kinematic equation we have that

$v^2 = u^2 +2as$

=> $a = \frac{v^2 - u^2 }{ 2(s_2 - s_1 )}$

=> $a = \frac{0 - 28^2 }{ 2(88 - 0 )}$

=> $a = - 4.45 m/s^2$

The negative sign shows that it is decelerating

6 0

4 years ago

When a Lunar Module landed on the Moon, it used thrusters to slow its descent to the surface. When other spacecraft are returned

katen-ka-za [31]

They had to use thrust because there is no air on the moon meaning that parachutes would be totally useless because parachutes require air to create drag. Fun fact since there is no air on the lunar surface sound waves cannot be made since there are no air molecules for the waves to pass through, so that's why they use radio's in space.

3 0

3 years ago