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3 years ago

A solution is oversaturated with solute. which could be done to decrease the oversaturation?

Physics

2 answers:

Rzqust [24]3 years ago

5 0

Answer: The correct awnser is B add more solute

Explanation:

Grace [21]3 years ago

4 0

<span>A solution is oversaturated with solute. The thing that could be done to decrease the oversaturation is to add more solvent in order to decrease the concentration of the solute. You can also increase the temperature to increase solubility of the solute. Hope this answers the question.</span>

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The medium through which a mechanical wave passes can be a solid, liquid, or gas. Properties of a wave change when it moves from

Tpy6a [65]

Answer:

Option A

Explanation:

Mechanical waves requires some medium to travel through. They travel faster in the dense medium as compared to a free medium.

The speed of a mechanical wave is fastest in the solid medium and the slowest in the gaseous medium. Hence, as the wave traverses from gaseous medium to the solid medium, its speed increases.

Thus, option A is correct

3 0

3 years ago

A jogger runs at a constant rate of 10.0 m every 2.0 seconds. The jogger starts at the origin and runs in the positive direction

Elis [28]

Answer:

(a) 25 m

(b) 75 m

Explanation:

Given that the jogger runs at a constant rate of 10.0 m every 2.0 seconds.

So, the speed of the jogger,

$v=\frac{10}{2}=5m/s\;\cdots(i)$

Let d be the distance covered by him in time, t s.

As distance=(speed) x (time)

So, $d=vt$

From equation (i)

$\Rightarrow d=5t\;\cdots(ii)$

As the jogger starts from origin, so, the distance, $d$ , also represents the position of the jogger at the time $t$ s.

The position-time graph has been shown.

(a) From equation (ii), for t=5.0 s

$d=5\times 5=25 m$

So, the jogger is at a distance of 25 m from the origin.

(b) Similarly, for t=15.0 s

$d=5\times 15=75 m$

So, the jogger is at a distance of 75 m from the origin.

8 0

3 years ago

is it true that the composition of the atmosphere changes every few kilometers as you move away from earth

laiz [17]

Yes,it's true ok? So how have you been doing

5 0

3 years ago

A 10n force is applied to a 25kg mass to slide it across a frictional surface. What is the acceleration of the mass?

klasskru [66]

Answer: a = 0.4m/s^2 - 9.8*c where c is the coefficient of kinetic friction of the surface

Explanation: We know that, by the second Newton's law, a = F/m

where a is the acceleration, F is the net force and m is the mass of the object.

Then, if the surface is frictionless, the total force applied in the object is 10N, and the mass of the object is 25kg, so the acceleration is:

a =10N/25kg = 0.4m/s^2.

But if the surface is frictional, there will be a force of friction applied in the mass (this depends on the coefficient of friction and the weight of the mass), this means that the acceleration will be reduced.

If = -(9.8*25)*c

where c is a number that is bigger than 0 and smaller than 1, is called the coefficient of kinetic friction.

So the total force is now:

F = (10 - 9.8*25*c)

Then, the acceleration in a frictional surface is equal to:

a = (10 - 9.8*25*c)/25 = 0.4m/s^2 - 9.8*c

6 0

3 years ago

Read 2 more answers

Two objects with a mass of 10,000-kg each and a distance of 1 meter between them.

Fofino [41]

Answer:

d) What is the force if we doubled both the masses AND we doubled the distance

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3 years ago