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OLga [1]
3 years ago
11

Why do you think the temperature does not change much during a phase change? If possible, discuss your answer with your classmat

es and teacher.
Physics
1 answer:
skelet666 [1.2K]3 years ago
6 0

Answer:

Explanation:

The fact is that there exists no temperature change until a phase change is complete. That is to say, during a phase change, the energy that will be supplied is used only to separate the molecules. There is no part of it is used to increase the kinetic energy of the molecules. So its temperature will not rise, not really, since the kinetic energy of the molecules remains the same. Also, it should be noted that the average kinetic energy of the molecules does not change at the moment that melting occurs, thus the temperature of the molecules does not change.

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The key to making a concise mathematical definition of escape velocity is to consider the energy. If an object is launched at it
aleksklad [387]

Answer:

The total Mechanical energy will be zero

Explanation: Escape velocity is the velocity required by a free object in order to overcome the impact of the force of gravity. The total mechanical energy of an object is the total energy possessed by an object which includes its kinectic and potential energy.

since the object is moving at an escape velocity which is 11.2m/s the object will be assumed to be weightless

Etotal = kinetic energy + potential energy

kinetic energy= 1/2*M*V*V

Potential energy=MGH

Etotal=1/2*0*11.2*11.2+0*0*0

Etotal=0+0

Etotal=0.

3 0
3 years ago
You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=
velikii [3]

Answer:

Approximately 4.2\; {\rm s} (assuming that the projectile was launched at angle of 35^{\circ} above the horizon.)

Explanation:

Initial vertical component of velocity:

\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}.

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing y_{1} is the same as the altitude y_{0} at which this projectile was launched: y_{0} = y_{1}.

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is 20.6\; {\rm m\cdot s^{-1}} (upwards,) the vertical velocity right before landing would be (-20.6\; {\rm m\cdot s^{-1}}) (downwards.) The change in vertical velocity is:

\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}.

Since there is no drag on this projectile, the vertical acceleration of this projectile would be g. In other words, a = g = -9.81\; {\rm m\cdot s^{-2}}.

Hence, the time it takes to achieve a (vertical) velocity change of \Delta v_{y} would be:

\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}.

Hence, this projectile would be in the air for approximately 4.2\; {\rm s}.

8 0
1 year ago
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Mr. Bennet's class completed an investigation on magnetism. They found that most metals were attracted to magnets and plastics w
Vinil7 [7]

Answer:

A

Explanation:

7 0
3 years ago
A pulley system lifts a 500-lb. block 2.0 feet with an effort of only 50 lb. If the 50 lb. moves 30 feet, calculate the efficien
tatiyna
The answer to this is 67%
4 0
3 years ago
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Convertir 1200 ms a cs<br> Convertir 0,3 mm a cm.
Vaselesa [24]

Answer:

You can do the reverse unit conversion from cm/s to m/s, or enter any two units below: Metre per second (U.S. spelling: meter per second) is an SI derived unit of both speed (scalar) and velocity (vector quantity which specifies both magnitude and a specific direction), defined by distance in metres divided by time in seconds.

Explanation:

3 0
3 years ago
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