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marusya05 [52]
3 years ago
7

4. What happens to energy during the formation of a solution?

Physics
1 answer:
Svetlanka [38]3 years ago
6 0

D It is released or absorbed.

Explanation:

To form a solution, energy is absorbed or released in the process.

A solution is a substance made up of solute dissolved in a solvent. The solute is the dissolved substance where as the solvent is the dissolving medium.

  • To form a solution, solutes are dissolved.
  • In this dissolution reaction, energy is used to break bonds within the solute to bring them in contact with the solvent.
  • The combination of the solute with the solvent is a bond formation process.
  • Bond formation is an exothermic process in which energy is released.
  • Bond breaking process is endothermic in which energy is absorbed.
  • To form solutions, bonds are formed and broken.

Learn more:

Solutions brainly.com/question/8781174

#learnwithBrainly

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Answer:

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Explanation:

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What will be the pressure exerterd by the 0bject if 4000n is acting on an area of 50msqure
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<h3>Given, </h3>

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You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=
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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}.

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