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Archy [21]
3 years ago
8

What does density have to do with heat?

Physics
2 answers:
joja [24]3 years ago
8 0
If you take a fluid (i.e. air or water) and heat it, the portion that is heated usually expands. The same mass takes up more volume and as a consequence the heated portion becomes less dense than the portion that is<span><span> not heated.</span> </span>
Vadim26 [7]3 years ago
7 0
Cooling a substance causes molecules to slow down and get<span> slightly closer together, occupying a smaller volume that results in an increase in </span>density<span>. Hot water is less dense and </span>will<span> float on room-temperature water. Cold water is more dense and </span>will<span>sink in room-temperature water.</span>
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c

Explanation:

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Elements with atomic numbers 58 through 71 are part of what series?
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Elements with atomic numbers from 58 through 71 are part of the

<span>lanthanide</span> series <span />

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A 45.0-N force pushes a cart 12.5 meters down a hallway. What is the work done on the cart?
makvit [3.9K]

Answer:

562.5J

Explanation:

The following were obtained from the question:

F = 45N

d = 12.5m

w =?

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3 years ago
What is a milliliter.
zavuch27 [327]

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one thousandth of a liter (0.002 pint).

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A person with mass mp = 74 kg stands on a spinning platform disk with a radius of R = 2.31 m and mass md = 183 kg. The disk is i
timurjin [86]

Answer:

1) 883 kgm2

2) 532 kgm2

3) 2.99 rad/s

4) 944 J

5) 6.87 m/s2

6) 1.8 rad/s

Explanation:

1)Suppose the spinning platform disk is solid with a uniform distributed mass. Then its moments of inertia is:

I_d = m_dR^2/2 = 183*2.31^2/2 = 488 kgm^2

If we treat the person as a point mass, then the total moment of inertia of the system about the center of the disk when the person stands on the rim of the disk:

I_{rim} = I_d + m_pR^2 = 488 + 74*2.31^2 = 883 kgm^2

2) Similarly, he total moment of inertia of the system about the center of the disk when the person stands at the final location 2/3 of the way toward the center of the disk (1/3 of the radius from the center):

I_{R/3} = I_d + m_p(R/3)^2 = 488 + 74*(2.31/3)^2 = 532 kgm^2

3) Since there's no external force, we can apply the law of momentum conservation to calculate the angular velocity at R/3 from the center:

I_{rim}\omega_{rim} = I_{R/3}\omega_{R/3}

\omega_{R/3} = \frac{I_{rim}\omega_{rim}}{I_{R/3}}

\omega_{R/3} = \frac{883*1.8}{532} = 2.99 rad/s

4)Kinetic energy before:

E_{rim} = I_{rim}\omega_{rim}^2/2 = 883*1.8^2/2 = 1430 J

Kinetic energy after:

E_{R/3} = I_{R/3}\omega_{R/3}^2/2 = 532*2.99^2/2 = 2374 J

So the change in kinetic energy is: 2374 - 1430 = 944 J

5) a_c = \omega_{R/3}^2(R/3) = 2.99^2*(2.31/3) = 6.87 m/s^2

6) If the person now walks back to the rim of the disk, then his final angular speed would be back to the original, which is 1.8 rad/s due to conservation of angular momentum.

3 0
3 years ago
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