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Mnenie [13.5K]
3 years ago
12

When a solid is in the process of changing into a liquid and extra energy is added to the system, the temperature does not chang

e.
What happens to this extra energy?
___________________________

A. The bonds between molecules are unaffected because the average kinetic energy does not change. Therefore, the molecular motion must increase.

B. The molecular motion of the water must increase while energy in the bonds decreases in order to keep the average kinetic energy the same.

C. The molecular motion does not change because the average kinetic energy does not change. The bonds are unaffected, because bonds do not change when going from solids to liquids. The energy must be needed somewhere else.

D. The temperature does not change during a phase change because the average kinetic energy does not change. Therefore, the potential energy in the bonds between molecules must change.
Physics
2 answers:
Lelu [443]3 years ago
8 0

Answer:

D. The temperature does not change during a phase change because the average kinetic energy does not change. Therefore, the potential energy in the bonds between molecules must change.

Explanation:

Fusion is the change from a solid to a liquid state. It occurs when a body, subjected to a given pressure, receives heat and its temperature reaches a certain value, at that moment the temperature does not change, during the phase change because the average kinetic energy does not change. Therefore, the potential energy in the bonds between molecules must change.

The amount of heat that the body must receive in order to become totally liquid, depends on the substance that constitutes it.

In general, when a substance is in the solid state, it has a well-defined shape. Its atoms are organized neatly in a structure called a crystalline network.

When it receives heat, the atoms that form the solid increase its vibration, increasing its temperature.

If the energy received increases, the vibration of the atoms will undo the crystalline network and the body will go into a liquid state.

Alekssandra [29.7K]3 years ago
7 0

Answer:

D. The temperature does not change during a phase change because the average kinetic energy does not change. Therefore, the potential energy in the bonds between molecules must change.

Explanation:

When there is a change of state (for example, from solid into a liquid, as in this example), when energy is added to the system, the temperature of the substance does not change.

The reason for this is that the energy supplied is no longer used to increase the average kinetic energy of the particle, but instead it is used to break the bonds between the different particles/molecules. For instance, since in this case the substance is changing from solid to liquid, all the energy supplied during the phase change is used to break the bonds between the molecules of the solid: when the process is done, all the molecules will be free to slide past each other, and the substance has turned completely into a liquid.

The bonds between molecules store potential energy: therefore, this means that the energy supplied during the phase change is not used to change the kinetic energy, but to change the potential energy in the bonds between the molecules.

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A mortar is like a small cannon that launches shells at steep angles. A mortar crew is positioned near the top of a steep hill.
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1) Distance down the hill: 1752 ft (534 m)

2) Time of flight of the shell: 12.9 s

3) Final speed: 326.8 ft/s (99.6 m/s)

Explanation:

1)

The motion of the shell is a projectile motion, so we  can analyze separately its vertical motion and its horizontal motion.

The vertical motion of the shell is a uniformly accelerated motion, so the vertical position is given by the following equation:

y=(u sin \theta)t-\frac{1}{2}gt^2 (1)

where:

u sin \theta is the initial vertical velocity of the shell, with u=156 ft/s and \theta=49.0^{\circ}

g=32 ft/s^2 is the acceleration of gravity

At the same time, the horizontal motion of the shell is a uniform motion, so the horizontal position of the shell at time t is given by the equation

x=(ucos \theta)t

where u cos \theta is the initial horizontal velocity of the shell.

We can re-write this last equation as

t=\frac{x}{u cos \theta} (1b)

And substituting into (1),

y=xtan\theta -\frac{1}{2}gt^2 (2)

where we have choosen the top of the hill (starting position of the shell) as origin (0,0).

We also know that the hill goes down with a slope of \alpha=-41.0^{\circ} from the horizontal, so we can write the position (x,y) of the hill as

y=x tan \alpha (3)

Therefore, the shell hits the slope of the hill when they have same x and y coordinates, so when (2)=(3):

xtan\alpha = xtan \theta - \frac{1}{2}gt^2

Substituting (1b) into this equation,

xtan \alpha = x tan \theta - \frac{1}{2}g(\frac{x}{ucos \theta})^2\\x (tan \theta - tan \alpha)-\frac{g}{2u^2 cos^2 \theta} x^2=0\\x(tan \theta - tan \alpha-\frac{gx}{2u^2 cos^2 \theta})=0

Which has 2 solutions:

x = 0 (origin)

and

tan \theta - tan \alpha=\frac{gx}{2u^2 cos^2 \theta}=0\\x=(tan \theta - tan \alpha) \frac{2u^2 cos^2\theta}{g}=1322 ft

So, the distance d down the hill at which the shell strikes the hill is

d=\frac{x}{cos \alpha}=\frac{1322}{cos(-41.0^{\circ})}=1752 ft=534 m

2)

In order to find how long the mortar shell remain in the air, we can use the equation:

t=\frac{x}{u cos \theta}

where:

x = 1322 ft is the final position of the shell when it strikes the hill

u=156 ft/s is the initial velocity of the shell

\theta=49.0^{\circ} is the angle of projection of the shell

Substituting these values into the equation, we find the time of flight of the shell:

t=\frac{1322}{(156)(cos 49^{\circ})}=12.9 s

3)

In order to find the final speed of the shell, we have to compute its horizontal and vertical velocity first.

The horizontal component of the velocity is constant and it is

v_x = u cos \theta =(156)(cos 49^{\circ})=102.3 ft/s

Instead, the vertical component of the velocity is given by

v_y=usin \theta -gt

And substituting at t = 12.9 s (time at which the shell strikes the hill),

v_y=(156)(cos 49^{\circ})-(32)(12.9)=-310.4ft/s

Therefore, the  final speed of the shell is:

v=\sqrt{v_x^2+v_y^2}=\sqrt{(102.3)^2+(-310.4)^2}=326.8 ft/s=99.6 m/s

Learn more about projectile motion:

brainly.com/question/8751410

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