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4 years ago

What’s the stored energy related to positions of atoms within molecules

Physics

1 answer:

zepelin [54]4 years ago

8 0

Cheimcal engery is related to positions of atoms within molecules..

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factors that affect health, including genetic, lifestyle, socioeconomic, and physical environmental factors.

scZoUnD [109]

Yes? I think a tutor would help with this!!!!

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3 years ago

In (A) Water has same mass and great volume

aniked [119]

Answer:

I didn't get pic from you but I got the type of answer that u want.

Explanation:

Water molecules are always moving. But on the average they are packed the same throughout. Therefore, the ratio between the mass and volume is the same, making the density the same. This is true no matter the size of the sample or where you select your sample from.

7 0

3 years ago

A red pool ball is rolling directly east before it collides with the

ikadub [295]

Answer: North East

Explanation: Trust me, I was just doing this on the Ck-12 and this the answer I choose and It said I'm correct.

7 0

3 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

at a constant pressure at a temperature of 280 k the gas in the cylinder has the volume of 10 liters if the volume of the gas is

Sav [38]

Since there is constant pressure, you can use Charles's Law:

V1/T1 = V2/T2

10L/280K = 20L/T

0.0357 = 20/ T

T = 20/0.0357

T = 560K which is the new temperature

7 0

3 years ago