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3 years ago

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Physics

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alina1380 [7]3 years ago

8 0

Answer:

We mentioned in the study section of Lecture 2 that hydrogen and oxygen combine in the ratio of 1 to 8, but that this is not enough information for leading to the conclusion that two hydrogen atoms combine with one of oxygen to form a water molecule. A key idea is attributed to Avagadro who said that equal volumes of gas (at the same temperature and pressure) contain equal numbers of constituent atoms or molecules. Experiments show that two liters of hydrogen gas will combine with one liter of oxygen gas to form two liters of water vapor. Each hydrogen molecule in hydrogen gas consists of two hydrogen atoms bonded together. Likewise, two oxygen atoms bind to make a oxygen molecule.

A "model" of a physical process is used to represent what one actually observes, even though this is an "ideal" model and not expected to be correct in all respects. However, it is a good enough model to explain many of the properties of gases with sufficient accuracy.

The motion of gas particles can be used to explain the pressure exerted and the temperature of a gas. The pressure on a surface is due to the force on that surface divided by its area. The force comes about from the multiple impacts of individual gas particles. Temperature, on the other hand, is DEFINED in terms of the average kinetic energy assocated with the motion of the gas particles. The greater the kinetic energy, the greater the temperature. See the apparatus shown in Figure 7.6 of the text which gives a simple way of measuring the distributions of speeds of atomic particles.

To visualize how gas particles colliding with a container create pressure, see Website II.

Gas particles move in all possible directions with differing speeds. The Kinetic Energy (KE) of a gas particle is equal to 1/2 its mass times its speeds squared. That is KE = 1/2 M x V2 , where M is the mass of the gas particle and V is its speed. The gas particles have a range of speeds, just like cars on a road, but it is the average of the speed squared times the mass, or the average kinetic energy which characterizes the temperature of a gas.

High temperature is associated with high kinetic energies and low temperatures are associated with low kinetic energies. However, keep in mind that the kinetic energy, and in this case the temperature, is proportional to the mass times the speed squared. So heavy particles moving more slowly will have the same kinetic energy as light particles moving more rapidly. Also, because the kinetic energy varies as the square of the speed, if two particles have the same mass, but one moves twice as fast as the other, it will have four times the kinetic energy (or temperature).

If temperature is associated with kinetic energy of a gas, one could ask at this point what controls the temperature of solids and liquids. It turns out that it is the kinetic energy of the constituent atoms and molecules that characterize the temperature of liquids and solids as well. We show in class a transparency picturing a solid with its atoms rigidly connected to each other. We will discuss more about liquids and solids in the next lecture, based on chapter 8. However, for now, let's keep in mind that the atoms or molecules in a solid, although bound to its neighbors in a rigid structure, can oscillate back and forth, and it is this motion that characterizes the temperature of a solid (or in a similar manner, of a liquid as well). As before, rapid oscillations mean high temperatures, and slower oscillations are lower temperatures.

4 - The Three Temperature Scales

There are three temperature scales. In the United States, we commonly use the Farenheit scale while in most other nations, the Celsius or Centigrade scale is used. Figure 7.10 shows these two scales side by side. Water boils at 212 degrees Farenheit or 100 degrees Centigrade. Water freezes at 32 degrees Farenheit or zero degrees Centigrade. However, the most important temperature scale for scientific calculations is the absolute temperature scale, or the Kelvin scale. Zero degrees Kelvin is the coldest possible temperature: it can be physically interpreted as the situation where the atoms or molecules have zero kinetic energy...so this is a very natural temperature scale. Zero degrees Kelvin is also -273 degrees Centigrade. Water freezes at +273 degrees Kelvin and zero degrees Centigrate. Hence, a difference of one degree is the same on the Centigrade and Kelvin scales, but the zero points are different.

R.S. Panvini

9/2/2002Explanation:

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This is greater than the initial charge, which violates the principle that the charge cannot be created or destroyed, consequently this distribution is impossible to achieve

Explanation:

The metals distribute the charge on all surface when they touch the surface increases so that charge density decreases and when the charge is separated into smaller in each metal.

Let's apply this principle to our case.

One of the spheres is loaded with a charge q, when touching a ball its charge is reduced to 1 / 2q for each ball.

qA = ½ q

qB = ½ q

qC = 0

The total charge is q

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If we touch the ball A again with the other sphere not charged C, the chare is distributed and when separated it is reduced by half

qA = 1/2 (q / 2) = ¼ q

qC = ¼ q

qB = ½ q

At this point all spheres have a charge,

qA = ¼ q

qb = ½ q

qC = ¼ q

The total charge is q

Now let's contact spheres B and one of the other two

Q = ½ q + ¼ q = ¾ q

When splitting the charge

qB = ½ ¾ q = 3/8 q

qC = ½ ¾ q = 3/8 q

qA = ¼ q

The total charge is q

Note that the total load is always equal to q

Now let's analyze the given configuration

Let's look for the total load

Q = qA + QB + QC

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A 4.87-kg ball of clay is thrown downward from a height of 3.21 m with a speed of 5.21 m/s onto a spring with k = 1570 N/m. The

Yuki888 [10]

Answer:

Approximately $0.560\; {\rm m}$ , assuming that:

the height of $3.21\; {\rm m}$ refers to the distance between the clay and the top of the uncompressed spring.
air resistance on the clay sphere is negligible,
the gravitational field strength is $g = 9.81\; {\rm m\cdot s^{-2}}$ , and
the clay sphere did not deform.

Explanation:

Notations:

Let $k$ denote the spring constant of the spring.
Let $m$ denote the mass of the clay sphere.
Let $v$ denote the initial speed of the spring.
Let $g$ denote the gravitational field strength.
Let $h$ denote the initial vertical distance between the clay and the top of the uncompressed spring.

Let $x$ denote the maximum compression of the spring- the only unknown quantity in this question.

After being compressed by a displacement of $x$ , the elastic potential energy $\text{PE}_{\text{spring}}$ in this spring would be:

$\displaystyle \text{PE}_{\text{spring}} = \frac{1}{2}\, k\, x^{2}$ .

The initial kinetic energy $\text{KE}$ of the clay sphere was:

$\displaystyle \text{KE} = \frac{1}{2}\, m \, v^{2}$ .

When the spring is at the maximum compression:

The clay sphere would be right on top of the spring.
The top of the spring would be below the original position (when the spring was uncompressed) by $x$ .
The initial position of the clay sphere, however, is above the original position of the top of the spring by $h = 3.21\; {\rm m}$ .

Thus, the initial position of the clay sphere ( $h = 3.21\; {\rm m}$ above the top of the uncompressed spring) would be above the max-compression position of the clay sphere by $(h + x)$ .

The gravitational potential energy involved would be:

$\text{GPE} = m\, g\, (h + x)$ .

No mechanical energy would be lost under the assumptions listed above. Thus:

$\text{PE}_\text{spring} = \text{KE} + \text{GPE}$ .

$\displaystyle \frac{1}{2}\, k\, x^{2} = \frac{1}{2}\, m\, v^{2} + m\, g\, (h + x)$ .

Rearrange this equation to obtain a quadratic equation about the only unknown, $x$ :

$\displaystyle \frac{1}{2}\, k\, x^{2} - m\, g\, x - \left[\left(\frac{1}{2}\, m\, v^{2}\right)+ (m\, g\, h)\right] = 0$ .

Substitute in $k = 1570\; {\rm N \cdot m^{-1}}$ , $m = 4.87\; {\rm kg}$ , $v = 5.21\; {\rm m\cdot s^{-1}}$ , $g = 9.81\; {\rm m \cdot s^{-2}}$ , and $h = 3.21\; {\rm m}$ . Let the unit of $x$ be meters.

$785\, x^{2} - 47.775\, x - 219.453 \approx 0$ (Rounded. The unit of both sides of this equation is joules.)

Solve using the quadratic formula given that $x \ge 0$ :

$\begin{aligned}x &\approx \frac{-(-47.775) + \sqrt{(-47.775)^{2} - 4 \times 785 \times (-219.453)}}{2 \times 785} \\ &\approx 0.560\; {\rm m}\end{aligned}$ .

(The other root is negative and is thus invalid.)

Hence, the maximum compression of this spring would be approximately $0.560\; {\rm m}$ .

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3 years ago