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3 years ago

What is the difference between the states of phase equilibrium and metastability?

Physics

1 answer:

erma4kov [3.2K]3 years ago

4 0

Answer:

The main difference is that the metastable state is not a state of equilibrium, but a state of non-equilibrium that is maintained for a long time. A metastable state is when the system approaches equilibrium in a very slow manner.

And on the other hand the phase equilibrium as the name says is a state of equilibrium in which there are more than two phases coexisting.

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The diagram shows a ramp with a toy car at the bottom. A string attached to the front of the car and the string goes over a pull

Annette [7]

The answer would be Gravity.

Gravity is pulling the weight down, which is pulling the car up the ramp.

3 0

3 years ago

on the surface of jupiter the acceleration due to gravity is about 3 times that on earth. how much would a 100kg rock weigh on j

madam [21]

Answer:

Weight on Jupiter will be equal to 2940 N

Explanation:

We have given given acceleration due to gravity on Jupiter is 3 times of acceleration due to gravity on earth

Acceleration due to gravity on earth $g=9.8m/sec^2$

So acceleration due to gravity on Jupiter = $g'=3\times 9.8=29.4m/sec^2$

Mass is given m = 100 kg

We have to find the weight

Weight is equal to W = mg, here m is mass and a is acceleration

So weight $W=100\times 29.4=2940N$

4 0

3 years ago

Read 2 more answers

A series L-R-C circuit consists of a 226 Ω resistor, a 27.4 mH inductor, a 11.55 µF capacitor, and an AC source of amplitude 15

DanielleElmas [232]

Answer: 363 Ω.

Explanation:

In a series AC circuit excited by a sinusoidal voltage source, the magnitude of the impedance is found to be as follows:

Z = √((R^2 )+〖(XL-XC)〗^2) (1)

In order to find the values for the inductive and capacitive reactances, as they depend on the frequency, we need first to find the voltage source frequency.

We are told that it has been set to 5.6 times the resonance frequency.

At resonance, the inductive and capacitive reactances are equal each other in magnitude, so from this relationship, we can find out the resonance frequency fo as follows:

fo = 1/2π√LC = 286 Hz

So, we find f to be as follows:

f = 1,600 Hz

Replacing in the value of XL and Xc in (1), we can find the magnitude of the impedance Z at this frequency, as follows:

Z = 363 Ω

6 0

3 years ago

A solid circular shaft and a tubular shaft, both with the same outer radius of c=co = 0.550 in , are being considered for a part

Norma-Jean [14]

Answer:

The power for circular shaft is 7.315 hp and tubular shaft is 6.667 hp

Explanation:

Polar moment of Inertia

$(I_p)s = \frac{\pi(0.55)4}2$

= 0.14374 in 4

Maximum sustainable torque on the solid circular shaft

$T_{max} = T_{allow} \frac{I_p}{r}$

= $(14 \times 10^3) \times (\frac{0.14374}{0.55})$

= 3658.836 lb.in

= $\frac{3658.836}{12}$ lb.ft

= 304.9 lb.ft

Maximum sustainable torque on the tubular shaft

$T_{max} = T_{allow}( \frac{Ip}{r})$

= $(14 \times10^3) \times ( \frac{0.13101}{0.55})$

= 3334.8 lb.in

= $(\frac{3334.8}{12} )$ lb.ft

= 277.9 lb.ft

Maximum sustainable power in the solid circular shaft

$P_{max} = 2 \pi f_T$

= $2\pi(2.1) \times 304.9$

= 4023.061 lb. ft/s

= $(\frac{4023.061}{550})$ hp

= 7.315 hp

Maximum sustainable power in the tubular shaft

$P _{max,t} = 2\pi f_T$

= $2\pi(2.1) \times 277.9$

= 3666.804 lb.ft /s

= $(\frac{3666.804}{550})$ hp

= 6.667 hp

7 0

3 years ago

You and a partner sit on the floor and stretch out a coiled spring to a length of 7.2 meters. You shake the coil so you

vekshin1

Answer:

Approximately $5.9\; {\rm m\cdot s^{-1}}$ (assuming that the partner is holding the other end of the coil stationary.)

Explanation:

In a standing wave, an antinode is a point that moves with maximal amplitude, while a node is a point that does not move at all. There is an antinode between every two adjacent nodes. Likewise, there is a node between every two adjacent antinodes.

The side of the spring that is being shaken moving with maximal amplitude. Hence, that point on this spring would also be an antinode. In contrast, the side of the spring that is held still (does not move at all) would be a node.

There would be a node between:

the antinode at the end of the spring that is being shaken, and
the antinode between the two ends of this spring.

Overall, the nodes and antinodes on this spring would be:

node at the end that is being held still,
antinode (as mentioned in the question),
node (inferred, not mentioned in the question), and
antinode at the end that is being shaken.

The distance between two adjacent nodes is equal to one-half (that is, $(1/2)$ ) the wavelength of the wave. The distance between a node and an adjacent antinode is one-quarter (that is, $(1/4)$ ) of the wavelength of the wave.

Thus, if the wavelength of the wave in this question is $\lambda$ , the length of this spring would be:

$\displaystyle \frac{1}{2}\, \lambda + \frac{1}{4}\, \lambda = \frac{3}{4}\, \lambda$ .

The question states that the length of this coiled spring is $7.2\; {\rm m}$ . In other words, $(3/4) \, \lambda = 7.2\; {\rm m}$ . The wavelength of this wave would be $(7.2\; {\rm m}) / (3/4) = 9.6\; {\rm m}$ .

The frequency $f$ of this wave is the number of cycles in unit time:

$\begin{aligned} f &= \frac{10}{16.3\; {\rm s}} \approx 0.613\; {\rm s^{-1}}\end{aligned}$ .

Hence, the speed $v$ of this wave would be:

$\begin{aligned} v &= \lambda\, f \\ &=9.6\; {\rm m} \times 0.613\; {\rm s^{-1}} \\ &\approx 5.9\; {\rm m \cdot s^{-1}}\end{aligned}$ .

3 0

2 years ago