Physics, 01.07.2019 22:00 sedratkawaiah13

The diagram shows the motion of a group of water molecules over a heat source. how do the molecules at point a compare to the molecules at point b? check all that apply. they are closer together. they are farther apart. they are moving faster. they are moving slower. they have gained energy. they have lost energy.

The diagram shows the motion of a group of water molecules over a heat source. how do the molecules

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Answer from: Lauraemia43

Point A is the point that represent the beginning of application of heat.
At point A, THE MOLECULES ARE CLOSER TOGETHER, THEY ARE MOVING SLOWER AND THEY HAVE LOST ENERGY.
Originally, the molecules of a liquid move slowly and are closer together before the application of heat or after the heat has been removed and they have cooled down.

Answer from: vince022503

The molecules at point A will be gain energy, move faster, and are farther apart.

The given picture shows the convection phenomena.

The molecules of point- A are heated by the heat source, so the kinetic energy of these particles will be high hence they move faster. Thus density will be low.

The molecules at point-B are not heated, so they have very low kinetic energy or move slower. The density will be high. Thus they sink.

Therefore, the molecules at point A will be gain energy, move faster, and are farther apart.

To know more about Convection:

link

Answer from: Quest

the small 0.38-kg slider is known to move from position a to position b along the vertical-plane slot. determine (a) the work done on the body by its weight and (b) the work done on the body by the spring. the distance r = 0.70 m, the spring modulus k = 154 n/m, and the unstretched length of the spring is 0.53 m

explanation:

-reverse card

Answer from: Quest

The figure is missing but we can answer anyway.the correct answer is "releasing the string". in fact, when the person is holding the bow and the arrow, the kinetic energy of the arrow is still zero (because the arrow is not moving). the moment the arrow is released, it acquires a velocity v, so its kinetic energy will become different from zero: $k= \frac{1}{2}mv^2$ where m is the mass of the arrow and v its velocity. all the other options, instead, do not increase the kinetic energy of the system: - pulling farther back on the string : this will just increase the elastic potential energy of the system- holding the arrow higher: this will increase the gravitational potential energy of the system (which is given by u=mgh, where h is the height above the ground)- decreasing the mass of the arrow: this will decrease the gravitational potential energy of the system (which is given by u=mgh, where m is the mass)

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