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3 years ago

13. Describe the molecules of a solid in terms of kinetic energy.

1 answer:

3 0

The kinetic molecular theory of matter states that: ... Molecules in the solid phase have the least amount of energy, while gas particles have the greatest amount of energy. The temperature of a substance is a measure of the average kinetic energy of the particles.

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In about 5 billion years, at the end of its lifetime, our sun will end up as a white dwarf, having about the same mass as it doe

vitfil [10]

It is fine to use the equation given by Plitter, since we are told that the mass is about the same as it is now, and I seriously doubt the original question wants the student to go into relativistic effects, electron degeneracy pressure and magnetic effects that govern a real white dwarf star.
There is no need to make it unnecessarily complicated, when the question is set at high school level. The question asks, given a particular radius, and a given mass, what will the density be (which in this case will be the average density). To answer the question, one needs to know the mass of the sun (which is about 2×1030 Kg. One needs to convert the diameter to a radius, and then calculate the spherical volume of the white dwarf. Then one can use the formula given above, namely density=mass/volume

8 0

3 years ago

Find the required angular speed, ω, of an ultracentrifuge for the radial acceleration of a point 2.00 cm from the axis to equal

MAXImum [283]

Answer:

Angular velocity of an ultra centrifuge is 17146.42 rad/s.

Explanation:

Given that,

Acceleration of an ultra centrifuge, $a=6\times 10^5\ g=6\times 10^5\times 9.8=5.88\times 10^6\ rad/s2$

Distance from axis, r = 2 cm = 0.02 m

We need to find the angular speed. We know that the relation between angular speed and angular acceleration is given by :

$a=r\omega^2\\\\\omega=\sqrt{\dfrac{a}{r}} \\\\\omega=\sqrt{\dfrac{5.88\times 10^6}{0.02}} \\\\\omega=17146.42\ rad/s$

So, the angular velocity of an ultra centrifuge is 17146.42 rad/s.

3 0

4 years ago

An electric turntable 0.730 mm in diameter is rotating about a fixed axis with an initial angular velocity of 0.240 rev/srev/s a

Zolol [24]

Answer:

a) $\omega = 0.421\,\frac{rev}{s}$ , b) $\Delta \theta = 0.066\,rev$ , c) $v = 0.966\,\frac{mm}{s}$ , d) $a = 3.293\,\frac{mm}{s^{2}}$

Explanation:

a) The angular velocity of the turntable after $0.200\,s$ .

$\omega = \omega_{o} + \alpha\cdot \Delta t$

$\omega = 0.240\,\frac{rev}{s} + (0.906\,\frac{rev}{s^{2}} )\cdot (0.2\,s)$

$\omega = 0.421\,\frac{rev}{s}$

b) The change in angular position is:

$\Delta \theta = \omega_{o}\cdot t + \frac{1}{2} \cdot \alpha \cdot t^{2}$

$\Delta \theta = (0.240\,\frac{rev}{s} )\cdot (0.2\,s) + \frac{1}{2}\cdot (0.906\,\frac{rev}{s^{2}} )\cdot (0.2\,s)^{2}$

$\Delta \theta = 0.066\,rev$

c) The tangential speed of a point on the rim of the turn-table:

$v = r\cdot \omega$

$v = (0.365\times 10^{-3}\,m)\cdot (0.421\,\frac{rev}{s} )\cdot (\frac{2\pi\,rad}{1\,rev} )$

$v = 9.655\times 10^{-4}\,\frac{m}{s}$

$v = 0.966\,\frac{mm}{s}$

d) The tangential and normal components of the acceleration of the turn-table:

$a_{t} = (0.365\times 10^{-3}\,m)\cdot (0.906\,\frac{rev}{s^{2}})\cdot (\frac{2\pi\,rad}{1\,rev} )$

$a_{t} = 2.078\times 10^{-3}\,\frac{m}{s^{2}}$

$a_{t} = 2.078\,\frac{mm}{s}$

$a_{n} = (0.365\times 10^{-3}\,m)\cdot \left[(0.421\,\frac{rev}{s} )\cdot (\frac{2\pi\,rad}{1\,rev} )\right]^{2}$

$a_{n} = 2.554\times 10^{-3}\,\frac{m}{s^{2}}$

$a_{n} = 2.554\,\frac{mm}{s^{2}}$

The magnitude of the resultant acceleration is:

$a = \sqrt{(2.078\,\frac{mm}{s} )^{2}+(2.554\,\frac{mm}{s} )^{2}}$

$a = 3.293\,\frac{mm}{s^{2}}$

8 0

3 years ago

If two objects at different temperatures are in contact with each other, what happens to their temperatures?

Anton [14]

If u put a cold & hot together, the temp with start to even out to the same temp.

3 0

4 years ago

Read 2 more answers

A coil of 1000 turns of wire has a radius of 12 cm and carries a counterclockwise current of 15 A. If it is lying flat on the gr

Andrew [12]

Answer:

0.01663 Nm

Explanation:

N = 1000, r = 12 cm = 0.12 m, i = 15 A, B = 5.8 x 10^-5 T, θ = 25

Torque = N i A B Sinθ

Torque = 1000 x 15 x 3.14 x 0.12 x 0.12 x 5.8 x 10^-5 x Sin 25

Torque = 0.01663 Nm

3 0

4 years ago