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vladimir2022 [97]
3 years ago
13

Calculate the quantity of work, in joules, associated with the compression of a gas from 5.64 L to 3.35 L by a constant pressure

of 1.21 atm ...?
Physics
1 answer:
ElenaW [278]3 years ago
5 0
In thermodynamics<span>, </span>work<span> performed by a system is the energy transferred by the system to its surroundings. It is calculated by the integral of pressure and volume of the system. At constant pressure, the expression for work would be as follows:

W = P(V2-V1)
W = 1.21(3.35 -5.64) = -2.77 L-atm
W = -2.77 (101.33) = -280.78 J</span>
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A 2120 kg car traveling at 13.4 m/s collides with a 2810 kg car that is initally at rest at a stoplight. The cars stick together
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Answer:

The coefficient of friction between the cars and the road is 0.859.

Explanation:

The two cars collide each other inelastically, then we can determine the resulting velocity by the Principle of Momentum Conservation:

m_{A}\cdot v_{A} + m_{B}\cdot v_{B} = (m_{A} + m_{B})\cdot v (1)

Where:

m_{A}, m_{B} - Masses of the cars, in kilograms.

v_{A}, v_{B} - Initial velocities of the cars, in meters per second.

v - Velocity of the resulting system, in meters per second.

If we know that m_{A} = 2120\,kg, v_{A} = 13.4\,\frac{m}{s }, m_{B} = 2810\,kg and v_{B} = 0\,\frac{m}{s}, then the  velocity of the resulting system:

v = \frac{m_{A}\cdot v_{A}+m_{B}\cdot v_{B}}{m_{A}+m_{B}}

v = \frac{(2120\,kg)\cdot \left(13.4\,\frac{m}{s} \right)+(2810\,kg)\cdot \left(0\,\frac{m}{s} \right)}{2120\,kg + 2810\,kg}

v = 5.762\,\frac{m}{s}

By Principle of Energy Conservation and Work-Energy Theorem, we understand that the initial translational kinetic energy (K), in joules, is dissipated due to work done by friction (W_{f}), in joules, that is to say:

K = W_{f} (2)

\frac{1}{2}\cdot (m_{A}+m_{B})\cdot v^{2} = \mu\cdot (m_{A}+m_{B})\cdot g \cdot s

\frac{1}{2}\cdot v^{2} = \mu \cdot g\cdot s (2b)

Where:

\mu - Coefficient of friction, no unit.

g - Gravitational acceleration, in meters per square second.

s- Travelled distance, in meters.

If we know that v = 5.762\,\frac{m}{s}, g = 9.807\,\frac{m}{s^{2}} and s = 1.97\,m, then the coefficient of friction is:

\mu = \frac{v^{2}}{2\cdot g\cdot s}

\mu = \frac{\left(5.762\,\frac{m}{s} \right)^{2}}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (1.97\,m)}

\mu = 0.859

The coefficient of friction between the cars and the road is 0.859.

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Each of the rods depicted below were machined from same stock metal. They were originally machined to be the same length, but th
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The force required to extend a rod increases as the cross sectional area

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The rod that experiences the largest force is <u>rod B</u>

Reason:

The elongation of a rod by the application of a force is given by the

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From the above equation, we have that the elongation is inversely

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The rod that experiences the largest force is the rod with the largest cross

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Answer:

It is easier to stop the bicycle moving at a lower velocity because it will require a <em>smaller force</em> to stop it when compared to a bicycle with a higher velocity that needs a<em> bigger force.</em>

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The question above is related to "Newton's Law of Motion." According to the <em>Third Law of Motion</em>, whenever an object exerts a force on another object <em>(action force)</em>, an equal force is exerted against it. This force is of the same magnitude but opposite direction.

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