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Anuta_ua [19.1K]
3 years ago
10

For a quantity of gas at a constant temperature, the pressure P is inversely proportional to the volume V. What is the limit of

P as V approaches 0 from the right? Explain what this means in the context of the problem.
Physics
1 answer:
IceJOKER [234]3 years ago
7 0

Answer:

lim_{V \to 0^{+}} P = k\lim_{V \to 0^{+}} \frac{1}{V} =\infty

This limit is not defined.

Explanation:

We need to remember first this law

The Boyle's law states that under a constant temperature when the pressure is inversely proportion to the volume.

So that means: P \propto \frac{1}{V}

And we can put an equal if we do this:

P = \frac{k}{V} where k is the proportional constant.

For this case we want to find the following limit:

lim_{V \to 0^{+}} P = \lim_{V \to 0^{+}} \frac{k}{V}

And using properties of limits we have:

lim_{V \to 0^{+}} P = k\lim_{V \to 0^{+}} \frac{1}{V}

So for this case this limit tend to infinity since we are dividing a constant by a very low number positive near to 0.

So then we can conclude that:

lim_{V \to 0^{+}} P = k\lim_{V \to 0^{+}} \frac{1}{V} =\infty

This limit is not defined.

Interpretation: we are seeing that if the volume decrease considerable with the temperature constant by the inverse relation between P and V, the value of P increases to with no limit.

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Answer:

Exercise 1;

The centripetal acceleration is approximately 94.52 m/s²

Explanation:

1) The given parameters are;

The diameter of the circle = 8 cm = 0.08 m

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The centripetal acceleration = v²/r = 1.944444²/0.04 ≈ 94.52 m/s²

The centripetal acceleration ≈ 94.52 m/s²

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A body of mass 25kg, moving at 3 ms per second on a rough horizontal floor brought to rest after sliding through a distance of 2
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You have to solve this by using the equations of motion:
u=3
v=0
s=2.5
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Then using the equation:
F=ma
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Then use the formula:
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45=u(25g)
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Therefore, the coefficient of friction is 0.18

Hope that helps




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A 388 Hz tuning fork is resonating in a closed tube on a warm day when the speed of sound is 346 m/s. What is the length of the
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A

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After flying for 15 min in a wind blowing 42 km/h at an angle of 19° south of east, an airplane pilot is over a town that is 48
masha68 [24]

Answer:

The speed of the airplane relative to the air is 209.47km/hr

Explanation:

Whenever we are solving a physics problem, it's really useful to start by drawing a diagram of the problem (See picture attached). It will help us visualize the problem better.

Now, we know that the plane flew for an amount of time of 15 minutes. For our dimensions to be the same, we need to turn those 15min to hours, like this:

15min*\frac{1hr}{60min}=0.25hr

Once our time is rewritten as hours, we can now calculate the velocity towards north of the plane.

V=\frac{distance}{time}

the plane traveled a distance to the north of 48km so the velocity is:

V=\frac{48km}{0.25hr}

so

V=192km/hr j

Now, we can calculate the x and y-components of the velocity of the wind. The problem states that the wind is blowing at 42km/hr at an angle of 19° south of east, so the x and y-components of the velocity of the wind are:

V_{x}=42km/hr*cos(-19^{o} )=39.71 i

and

V_{y}=42km/hr*sin(-19^{o} )=-13.67 j

So the velocity of the wind can be expressed as a vector as:

V_{wind}=(39.71i - 13.67j)km/hr

Once we know this, we can find the velocity of the plane with respect of the wind on x and on y:

V_{plane x}=V_{plane/wind x}+V_{wind x}

V_{plane/wind x}=V_{plane x}-V_{wind x}

V_{plane/wind x}=(0-39.71 i)km/hr

V_{plane/wind x}= -39.71 i km/hr

and

V_{plane y}=V_{plane/wind y}+V_{wind y}

V_{plane/wind y}=V_{plane y}-V_{wind y}

V_{plane/wind y}=192km/hr j - (- 13.67j)km/hr

V_{plane/wind x}= 205.67 j km/hr

So the velocity of the plane with respect to the wind can be rewritten as:

V_{plane/wind x}= (-39.71i + 205.67 j) km/hr

Since the problem asks us to find the speed of the plane with respect to the wind, this means that we need to find the magnitude of the velocity, since the speed is a scalar defined to be the magnitude of the velocity.

so:

speed=\sqrt{(-39.71)^{2}+(205.67)^{2}  }

speed= 209.47 km/hr

Therefore, the speed of the airplane relative to the air is 209.47km/hr

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