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stiks02 [169]
3 years ago
13

Describe and explain how the movement and arrangement of the particles in a block of ice change as the ice melts to form liquid

water. In your answer, include ideas about energy transfer.
Physics
1 answer:
-BARSIC- [3]3 years ago
4 0

Answer:

Ice is water in solid phase, in this phase, the particles are very close together and relatively in fixed positions.

As the temperature starts to increase (thermal energy), also does the kinetic energy of the particles (so we have a change from thermal energy to kinetic energy), so they start to move "more", and the position of the particles starts to be less "fixed". There is a point where the particles have enough energy, and this point is where the phase of the water changes from solid to liquid phase (the fusion point). After this point the water can not hold his shape, and takes the shape of the container where it is.

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A car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of 1.90 m/s2
Ahat [919]

Answer:

Approximately 0.608 (assuming that g = 9.81\; \rm N\cdot kg^{-1}.)

Explanation:

The question provided very little information about this motion. Therefore, replace these quantities with letters. These unknown quantities should not appear in the conclusion if this question is actually solvable.

  • Let m represent the mass of this car.
  • Let r represent the radius of the circular track.

This answer will approach this question in two steps:

  • Step one: determine the centripetal force when the car is about to skid.
  • Step two: calculate the coefficient of static friction.

For simplicity, let a_{T} represent the tangential acceleration (1.90\; \rm m \cdot s^{-2}) of this car.

<h3>Centripetal Force when the car is about to skid</h3>

The question gave no information about the distance that the car has travelled before it skidded. However, information about the angular displacement is indeed available: the car travelled (without skidding) one-quarter of a circle, which corresponds to 90^\circ or \displaystyle \frac{\pi}{2} radians.

The angular acceleration of this car can be found as \displaystyle \alpha = \frac{a_{T}}{r}. (a_T is the tangential acceleration of the car, and r is the radius of this circular track.)

Consider the SUVAT equation that relates initial and final (tangential) velocity (u and v) to (tangential) acceleration a_{T} and displacement x:

v^2 - u^2 = 2\, a_{T}\cdot x.

The idea is to solve for the final angular velocity using the angular analogy of that equation:

\left(\omega(\text{final})\right)^2 - \left(\omega(\text{initial})\right)^2 = 2\, \alpha\, \theta.

In this equation, \theta represents angular displacement. For this motion in particular:

  • \omega(\text{initial}) = 0 since the car was initially not moving.
  • \theta = \displaystyle \frac{\pi}{2} since the car travelled one-quarter of the circle.

Solve this equation for \omega(\text{final}) in terms of a_T and r:

\begin{aligned}\omega(\text{final}) &= \sqrt{2\cdot \frac{a_T}{r} \cdot \frac{\pi}{2}} = \sqrt{\frac{\pi\, a_T}{r}}\end{aligned}.

Let m represent the mass of this car. The centripetal force at this moment would be:

\begin{aligned}F_C &= m\, \omega^2\, r \\ &=m\cdot \left(\frac{\pi\, a_T}{r}\right)\cdot r = \pi\, m\, a_T\end{aligned}.

<h3>Coefficient of static friction between the car and the track</h3>

Since the track is flat (not banked,) the only force on the car in the horizontal direction would be the static friction between the tires and the track. Also, the size of the normal force on the car should be equal to its weight, m\, g.

Note that even if the size of the normal force does not change, the size of the static friction between the surfaces can vary. However, when the car is just about to skid, the centripetal force at that very moment should be equal to the maximum static friction between these surfaces. It is the largest-possible static friction that depends on the coefficient of static friction.

Let \mu_s denote the coefficient of static friction. The size of the largest-possible static friction between the car and the track would be:

F(\text{static, max}) = \mu_s\, N = \mu_s\, m\, g.

The size of this force should be equal to that of the centripetal force when the car is about to skid:

\mu_s\, m\, g = \pi\, m\, a_{T}.

Solve this equation for \mu_s:

\mu_s = \displaystyle \frac{\pi\, a_T}{g}.

Indeed, the expression for \mu_s does not include any unknown letter. Let g = 9.81\; \rm N\cdot kg^{-1}. Evaluate this expression for a_T = 1.90\;\rm m \cdot s^{-2}:

\mu_s = \displaystyle \frac{\pi\, a_T}{g} \approx 0.608.

(Three significant figures.)

7 0
3 years ago
The cheetahs in Kenya and the cheetahs in Tanzania make up separate _____.
Andreyy89
C. Populations. 
  Hope that's right. 
6 0
3 years ago
Read 2 more answers
As you sit in a fishing boat, you noticed that 22 waves pass the boat every 60 seconds . If the distance from one crest to the n
Zinaida [17]

Answer:

0.1835m/s

Explanation:

The formula for calculating the speed of wave is expressed as;

v = fλ

f is the frequency - The number of oscillations completed in one seconds

If 22 waves pass the boat every 60 seconds,

number of wave that passes in 1 seconds = 22/60 = 0.367 waves

Therefore the frequency f of the wave is 0.367Hertz

λ (wavelength) is the distance between successive crest and trough of a wave

λ = 0.5m

Substitute the given values into the formula

v = fλ

v = 0.367 * 0.5

v = 0.1835

Hence the speed of the waves is 0.1835m/s

8 0
2 years ago
Explain how astronomers might use spectroscopy to determine the composition and temperature of a star.
Damm [24]

Everything starts from spectroscopy. Astronomers only have concentrated information at wavelengths that are emitted from the stars. What they do with this information is to obtain the frequency range of the stars and through spectroscopes they are responsible for dividing the radiation beams and determining the coincidence with the emission of those same waves, of chemical elements. From these observation techniques it is possible to obtain the composition and according to the color, obtaining characteristics such as temperature. The spectrum of stars consists of dark and bright lines called Fraunhofer lines. This spectrum is compared to the spectrum of different elements to find the composition of the stars. This is possible because the elements emit or absorb only specific wavelengths.

4 0
3 years ago
Which collects light in a reflecting telescope?
Alecsey [184]
Concave lens. These are used in making the objectives of reflection telescopes
4 0
3 years ago
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