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3 years ago

How does an increase in thermal energy affect the kinetic energy of the particles in each state of matter

Physics

1 answer:

Alenkinab [10]3 years ago

7 0

Explanation:

when there is an increase in thermal energy, the kinetic energy if the matter increases as well.

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Which of the following documents could be used to represent claims to financial assets?

stiks02 [169]

I think the correct answer would be passbooks. It can be used to prove one's financial assets. It holds record of all bank transactions done on that particular account. Basically, all withdrawals and deposits are recorded there. Hope this helped.

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4 years ago

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Einstein developed much of his understanding of relativity through the use of gedanken, or thought, experiments.

Zielflug [23.3K]

True. Thought experiments are what he called them, it’s like meditation

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3 years ago

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A solid sphere, solid cylinder, and a hollow pipe all have equal masses and radii. If the three of them are released simultaneou

Roman55 [17]

Answer:

The solid sphere will reach the bottom first.

Explanation:

In order to develop this problem and give it a correct solution, it is necessary to collect the concepts related to energy conservation. To apply this concept, we first highlight the importance of conserving energy so we will match the final and initial energies. Once this value has been obtained, we will concentrate on finding the speed, and solving what is related to the Inertia.

In this way we know that,

$\Delta KE = - \Delta PE$

$KE_t + KE_r = mgh$

We know as well that the lineal and angular energy are given by,

$KE_r = \frac{1}{2}I\omega^2$

And the tangential kinetic energy as

$KE_t = \frac{1}{2} mv^2$

Where $\omega = \frac{v}{R}$

Replacing

$\frac{1}{2}mv^2 + \frac{1}{2}I\frac{v}{R} = mgh$

Re-arrange for v,

$v=\sqrt{\frac{2mgh}{m+I/R^2}}$

We have here three different objects: solid cylinder, hollow pipe and solid sphere. We need the moment inertia of this objects and replace in the previous equation found, then,

For hollow pipe:

$I_{hp}=mR^2$

$v_{hp}=\sqrt{\frac{2mgh}{m+(mR^2)/R^2}}$

$v_{hp}=\sqrt{\frac{2mgh}{m+m)}$

$v_{hp}=\sqrt{gh}$

For solid cylinder:

$I_{sc}=\frac{1}{2}mR^2$

$v_{sc}=\sqrt{\frac{2mgh}{m+(1/2mR^2)/R^2}}$

$v_{sc}=\sqrt{\frac{2mgh}{m+1/2m}}$

$v_{sc}=\sqrt{\frac{3}{4}gh}$

For solid sphere,

$I_{ss}=\frac{2}{5}mR^2$

$v_{ss}=\sqrt{\frac{2mgh}{m+(2/5mR^2)/R^2}}$

$v_{ss}=\sqrt{\frac{2mgh}{m+2/5m}}$

$v_{ss}=\sqrt{\frac{10}{7}gh}$

Then comparing the speed of the three objects we have:

$v_{hp}$

$\sqrt{gh}$

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4 years ago

Parallel rays from a distant object are traveling in air and then are incident on the concave end of a glass rod with a radius o

skad [1K]

Answer:

the distance of image from the vertex is 45 cm and the image formed is in the glass.

Explanation:

distance of object, u = - infinity

radius of curvature, R = - 15 cm

refractive index, n = 1.5

Let the distance of image is v.

Use the formula

$-\frac{n1}{u}+\frac{n2}{v}=\frac{n2- n1}{R}\\\\-\frac{1}{\infty }+\frac{1.5}{v}=\frac{1.5-1}{-15}\\\\v=45 cm$

The image is in the glass.

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3 years ago

How are length and volume alike

SIZIF [17.4K]

You used lengths to calculate the volume.length is considered one dimensional.while volume is considered 3 dimensional.you used 3 individual dimensions their numerical lengths to find volumes of three dimensional figures.

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4 years ago

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