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3 years ago

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

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To reach its intended destination on time a cruise ship needs to travel north at 70 mph. However, it is travelling in the Gulf o

bogdanovich [222]

Answer:

speed is 81.03 mph

direction is N 3.58 W

Explanation:

given data

travel north = 70 mph

Stream current = 12 mph

direction = S 25° E

result due north = 70 mph

to find out

speed and direction

solution

we will get component of resultant that is

v cosθ and v sinθ

( 12cos295 , 12 sin295 ) at ( 0, 70)

as that we can say

v sinθ + 12sin295 = 70 ....................1

v cosθ + 12 cos295 = 0 ......................2

vcosθ = -5.0714

vsinθ = 80.8756

now by ratio

cosθ /sinθ = -5.0714/ 80.8756

cot θ = -0.0627

θ = 93.58

so direction is N 3.58 W

and

we know

vcosθ = - 12cos295

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4 0

3 years ago

As part of your daily workout, you lie on your back and push with your feet against a platform attached to two stiff springs arr

Nitella [24]

Answer:

1. Part A: 648N
2. Part B: 324J
3. Part C: 1,296N

Explanation:

1. Part A:

The magnitude of the force is calculated using the Hook's law:

$|F|=k\Delta x$

You know $\Delta x=0.250m$ but you do not have $k$ .

You can calculate it using the equation for the work-energy for a spring.

The work done to compress the springs a distance $\Delta x$ is:

$Work=\Delta PE=(1/2)k(\Delta x)^2$

Where $\Delta PE$ is the change in the elastic potential energy of the "spring".

Here you have two springs, but you can work as if they were one spring.

You know the work (81.0J) and the length the "spring" was compressed (0.250m). Thus, just substitute and solve for k:

$81.0J=(1/2)k(0.250m)^2\\\\k=2,592N/m$

In reallity, the constant of each spring is half of that, but it is not relevant for the calculations and you are safe by assuming that it is just one spring with that constant.

Now calculate the magnitude of the force:

$|F|=k\Delta x=2,592N/m\times 0.250m=648N$

2. Part B. How much additional work must you do to move the platform a distance 0.250 m farther?

The additional work will be the extra elastic potential energy that the springs earn.

You already know the elastic potential energy when Δx = 0.250m; now you must calculate the elastic potential energy when Δx = 0.250m + 0.250m = 0.500m.

$\Delta E=(1/2)2,592n/m\times(0.500m)^2=324J$

Therefore, you must do 324J of additional work to move the plattarform a distance 0.250 m farther.

3. Part C

What maximum force must you apply to move the platform to the position in Part B?

The maximum force is when the springs are compressed the maximum and that is 0.500m

Therefore, use Hook's law again, but now the compression length is Δx = 0.500m

$|F|=k\Delta x=2,592N/m\times 0.500m =1,296N$

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3 years ago

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Harlamova29_29 [7]

Answer:

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Explanation:

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3 years ago