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3 years ago

PLEASE HELP ASAP!!!

Physics

2 answers:

Tju [1.3M]3 years ago

4 0

Answer:

Heat is transferred to the bottom layer of water via conduction. Heat transfer via conduction takes place by vibration and collision of molecules.

The heated layer of water expands and rises. The cold top layer being dense rushes to the bottom to take its place.

Heat transfer in pot of water takes place via conduction and convection. The temperature of the water rises till whole water is uniformly heated.

joja [24]3 years ago

3 0

The behaviour of the water molecules are that they would vibrate faster and the space between them would also increase. Therefore, the result molecules' motion is that it will expand and take up more space. Temperature is affected by making it rise, when the temperature in the water rises, this can result to boiling, steam or vapour in which some of the water molecules will escape and evaporate into the atmosphere.

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A particle moves along a line so that its velocity at time t is v(t) = t2 − t − 20 (measured in meters per second). (a) Find the

Genrish500 [490]

Answer:

Displacement is 47,166667 meters, Distance traveled is 78.5 meters.

Explanation:

Let's remberi that $v(t) = \frac d{dt}s(t)$ . Distance traveled will be $\int\limits^{10}_1 v(t) \, dt$ , while displacement will be $\int\limits^{10}_1 |v(t)| \, dt$ (note the absolute value is needed because part of the movement MIGHT be backwards, which will add to the distance traveled, ie the number of steps, but does not contribute to displacement - think of a guy running in cirles).

Distance: $\int\limits^{10}_1 {t^2-t-20} dt =(\frac13 t^3 - \frac12 t^2 -20t )\limits^{10}_1 = (\frac 13 *1000 - \frac12*100-20*10) - (\frac13 -\frac 12 -20) = \frac{2000-300-1200-2+3+60}6 = \frac{561}6 =78.5$

Displacement: absolute value makes things a bit more annoying: v(t) is negative between $t=1$ and $t=5$ - the particle is moving backwards. The integral becomes:

$\int\limits^{10}_1 |v(t)| \, dt = -\int\limits^5_1t^2-t-20\,dt +\int\limits^{10}_5t^2-t-20\, dt =\\-(\frac13 t^3 -\frac12 t^2-20t)\limits^5_1 +(\frac13 t^3 -\frac12 t^2-20t)\limits^{10}_5 =\\ -(\frac13*125-\frac1/2*25-20*5)+(\frac13-\frac12-20)+(\frac13*1000-\frac12*100-20*10)-(\frac13*125-\frac1/2*25-20*5) = 289/6 = 48,166667$

Puff, pant, that should be it, please double check calculations, just in case.

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Learn more about action-reaction pairs at: brainly.com/question/12800382

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