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

luda_lava [24]

When the tractor pulls, the boy moves. When the elephant moves, the boy moves. If the elephant moves and has more power than the tractor, the tractor and the boy moves. If the tractor moves and has more power than the elephant, then the elephant and the boy moves. I hope this helps you, have a great day!

5 0

3 years ago

Read 2 more answers

HELP SCIENCE 8 15 pts

Bingel [31]

Answer:

scalar is having only magnitude, not direction.

Vector is a quantity having direction as well as magnitude, especially as determining the position of one point in space relative to another.

you find the differences of both of them i actually kinda gave it to you.

Explanation:

4 0

3 years ago

A dime is placed in front of a concave mirror that has a radius of curvature R = 0.40 m. The image of the dime is inverted and t

andrew11 [14]

Answer:

distance between the dime and the mirror, u = 0.30 m

Given:

Radius of curvature, r = 0.40 m

magnification, m = - 2 (since,inverted image)

Solution:

Focal length is half the radius of curvature, f = $\frac{r}{2}$

f = $\frac{0.40}{2} = 0.20 m$

Now,

m = - $\frac{v}{u}$

- 2 = - $\frac{v}{u}$

$\frac{v}{u}$ = 2 (2)

Now, by lens maker formula:

$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$

$\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$

v = $\frac{uf}{u - f}$ (3)

From eqn (2):

v = 2u

put v = 2u in eqn (3):

2u = $\frac{uf}{u - f}$

2 = $\frac{f}{u - f}$

2(u - 0.20) = 0.20

u = 0.30 m

6 0

3 years ago

A hollow cylinder of mass 2.00 kg, inner radius 0.100 m, and outer radius 0.200 m is free to rotate without friction around a ho

Aneli [31]

Answer:

h=2.86m

Explanation:

In order to give a quick response to this exercise we will use the equations of conservation of kinetic and potential energy, the equation is given by,

$\Delta PE_i + \Delta KE_i = \Delta PE_f +\Delta KE_f$

There is no kinetic energy in the initial state, nor potential energy in the end,

$mgh+0=0+KE_f$

In the final kinetic energy, the energy contributed by the Inertia must be considered, as well,

$mgh = (\frac{1}{2}mv^2+\frac{1}{2}I\omega^2)$

The inertia of the bodies is given by the equation,

$I=\frac{m(R_1^2+R^2_2)}{2}$

$I=\frac{2(0.2^2+0.1^2)}{2}$

$I=0.05Kgm^2$

On the other hand the angular velocity is given by

$\omega =\frac{v}{R_2}=\frac{4}{1/5} = 2rad/s$

Replacing these values in the equation,

$(0.5)(9.8)(h) =\frac{1}{2}*0.5*4^2+\frac{1}{2}*0.05*20^2$

Solving for h,

$h=2.86m$

5 0

3 years ago

In a drill during basketball practice, a player runs the length of the 30-meter court and back. The player does this three times

Sergeeva-Olga [200]

Answer:

0 m/s

Explanation:

Average velocity of an object is given by the net displacement divided by time taken. Displacement is equal to the shortest path covered by the object.

In this problem, a player runs the length of the 30-meter court and back. The player does this three times in 60 seconds.

As the player runs the court and returns to the original point. It would mean that the shortest path covered is 0.

Average velocity = displacement/time

v=0/30

v = 0 m/s

Hence, the correct option is (1).

6 0

4 years ago