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3 years ago

6

An automobile tire having a temperature of −1.5 ◦C (a cold tire on a cold day) is filled to a gauge pressure of 26 lb/in2 . What

would be the gauge pressure in the tire when its temperature rises to 40◦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 .

1 answer:

Salsk061 [2.6K]3 years ago

6 0

Answer: $32.22\ lb/in^2$

Explanation:

Given

Temperature on the hot day $T_i=-1.5^{\circ} C\approx 271.5\ K$

Gauge Pressure $P_i=26\ lb/in^2$

When Temperature rises to $T_f=40^{\circ}C\approx 313\ K$ on a hot day then

absolute pressure is let say $P_f$

Using ideal gas equation we can write

$\Rightarrow \frac{PV}{T}=constant$

as volume is constant therefore

$\Rightarrow \frac{P}{T}=constant$

$\Rightarrow \frac{26+14.7}{271.5}=\frac{P_f}{313}$

$\Rightarrow P_f=46.92\ lb/in^2$

Gauge Pressure is given by

$\Rightarrow 46.92-14.7=32.22\ lb/in^2$

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Answer:

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$F$ - Force, measured in newtons.

$k$ - Proportionality ratio, measured in newton-meters.

$r$ - Distance respect to axis of rotation passing through hinges, measured in meters.

From (Eq. 1) we get the following relationship and clear the final force within:

$F_{A}\cdot r_{A} = F_{B}\cdot r_{B}$

$F_{B}=\left(\frac{r_{A}}{r_{B}} \right)\cdot F_{A}$ (Eq. 2)

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$F_{A}$ , $F_{B}$ - Initial and final forces, measured in newtons.

$r_{A}$ , $r_{B}$ - Initial and final distances, measured in meters.

If we know that $F_{A} = 5\,N$ , $r_{A} = 0.9\,m$ and $r_{B} = 0.35\,m$ , then final force is:

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5)

In physics, forces are interactions that are able to change the velocity of an object.

Force is a vector quantity, so it has a magnitude and a direction.

The SI units of the force is the Newton (N).

Whenever an unbalanced force is applied to an object, the object experiences an acceleration, according to Newton's second law of motion:

$F=ma$

where

F is the force

m is the mass of the object

a is its acceleration

So, the acceleration of an object is proportional to the force applied:

$a=\frac{F}{m}$

6)

In physics, arrows are used to represent vector quantities. Therefore, they are also used to represent forces.

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- The length of the arrow is proportional to the magnitude of the vector quantity

- The direction of the arrow corresponds to the direction of the vector quantity

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7)

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F is the force

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- The acceleration is measured in meters per second squared ( $m/s^2$ )

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B)

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3)

Mass is a scalar quantity; it gives us a measure of the "amount of matter" contained in an object.

The SI unit of the mass is the kilogram (kg).

Being a scalar, mass has no direction, but only a magnitude.

Moreover, the mass is an intrinsec property of an object: therefore, it does not depend on the location of the object. So, an object has always the same mass, either it is on Earth or on another planet.

On the other hand, the force of gravity on an object depends on its location, so it changes.

4)

As we said in part 3), gravity is an attractive force that exists between all objects that have mass.

The magnitude of the force of gravity between two objects is given by the Universal Law of gravitation:

$F=\frac{Gm_1 m_2}{r^2}$

where

G is the gravitational constant

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r is the separation between the objects

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- all objects are attracted to one another with a gravitational force that is proportional to the mass of the objects and inversely proportional to the square of the distance between them.

And so:

a. When the mass of one or both objects increases, the gravitational force between the objects increases

b. When the distance between two objects increases, the attraction between the objects decreases

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