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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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A girl of mass m1=60 kilograms springs from a trampoline with an initial upward velocity of v1=8.0 meters per second. At height
AleksandrR [38]

a) 5.0 m/s

This first part of the problem can be solved by using the conservation of energy. In fact, the mechanical energy of the girl just after she jumps is equal to her kinetic energy:

E_i=\frac{1}{2}m_1v_1^2

where m1 = 60 kg is the girl's mass and v1 = 8.0 m/s is her initial velocity.

When she reaches the height of h = 2.0 m, her mechanical energy is sum of kinetic energy and potential energy:

E_f = \frac{1}{2}m_1 v_2 ^2 + m_1 gh

where v2 is the new speed of the girl (before grabbing the box), and h = 2.0m. Equalizing the two equations (because the mechanical energy is conserved), we find

\frac{1}{2}m_1 v_1^2 = \frac{1}{2}m_1 v_2 ^2 + m_1 gh\\v_1^2 = v_2^2 +2gh\\v_2 = \sqrt{v_1^2 -2gh}=\sqrt{(8.0 m/s)^2-(2)(9.8 m/s^2)(2.0 m)}=5.0 m/s

b) 4.0 m/s

After the girl grab the box, the total momentum of the system must be conserved. This means that the initial momentum of the girl must be equal to the total momentum of the girl+box after the girl catches the box:

p_i = p_f\\m_1 v_2 = (m_1 + m_2) v_3

where m2 = 15 kg is the mass of the box. Solving the equation for v3, the combined velocity of the girl+box, we find

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c) 2.8 m

We can use again the law of conservation of energy. The total mechanical energy of the girl after she catches the box is sum of kinetic energy and potential energy:

E_i = \frac{1}{2}(m_1+m_2) v_3^2 + (m_1+m_2)gh=\frac{1}{2}(75 kg)(4 m/s)^2+(75 kg)(9.8 m/s^2)(2.0m)=2070 J

While at the maximum height, the speed is zero, so all the mechanical energy is just potential energy:

E_f = (m_1 +m_2)gh_{max}

where h_max is the maximum height. Equalizing the two expressions (because the mechanical energy must be conserved) and solving for h_max, we find

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A displacement vector 5 units long makes an angle of 67° with the x-axis. What is the horizontal component of displacement?
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Yo sup??

magnitude of original vector is 5 units.

angle made with x axis is 67°

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