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3 years ago

How do salt and water form a solution?

Physics

2 answers:

iogann1982 [59]3 years ago

8 0

It would be D. The solute, salt, must dissolve IN the solvent, water (:

Klio2033 [76]3 years ago

5 0

<u>Answer:</u> The solute, salt must dissolve in the solvent, water.

<u>Explanation:</u>

Solute is defined as the substance which is present in smaller proportion in a solution.

Solvent is defined as the substance which is present in larger proportion in a solution.

Solution is the combination of solute and solvent.

In a solution, a solute will always dissolve in solvent.

We are given:

Salt and water to form a solution. Here, salt is present in smaller proportion, so it is a solute and water is a solvent.

Hence, the solute, salt must dissolve in the solvent, water.

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Answer:

S = V0 t + 1/2 a t^2

S = 5 m/s * 300 s + 1/2 * 1.2 m/s * (300 s^2)

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A constant force of 45 N directed at angle θ to the horizontal pulls a crate of weight 100 N from one end of a room to another a

Rasek [7]

Answer:

$W=173.48J$

Explanation:

information we know:

Total force: $F=45N$

Weight: $w=100N$

distance: $4m$

vertical component of the force: $F_{y}=12N$

-------------

In this case we need the formulas to calculate the components of the force (because to calculate the work we need the horizontal component of the force).

horizontal component: $F_{x}=Fcos\theta$

vertical component: $F_{y}=Fsen\theta$

but from the given information we know that $F_{y}=12N$

so, equation these two $F_{y}=Fsen\theta$ and $F_{y}=12N$

$Fsen\theta =12N$

and we know the force $F=45N$ , thus:

$45sen\theta=12$

now we clear for $\theta$

$sen\theta =12/45\\\theta=sin^{-1}(12/45)\\\theta =15.466$

the angle to the horizontal is 15.466°, with this information we can calculate the horizontal component of the force:

$F_{x}=Fcos\theta$

$F_{x}=45cos(15.466)\\F_{x}=43.37N$

whith this horizontal component we calculate the work to move the crate a distance of 4 m:

$W=F_{x}*D\\W=(43.37N)(4m)\\W=173.48J$

the work done is W=173.48J

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