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3 years ago

A solid is stirred into a liquid and dissolves. Which type of mixture forms?

Physics

2 answers:

hodyreva [135]3 years ago

7 0

Answer: The correct answer is solution.

Explanation:

A solution is a homogeneous mixture of two or more substance. The substance which is dissolved in another substance is called solute. The solvent is the liquid which dissolves in solid, liquid and gas.

For example, salt is put into water. It gets dissolved completely. Here, salt is solute and water is solvent.

In the given problem, a solid is stirred into a liquid and gets dissolve. A solid is dissolved into liquid.

Therefore, the type of mixture forms solution.

AVprozaik [17]3 years ago

5 0

The type of mixture that is formed when a solid is stirred into a liquid and dissolves is called suspension. The particles involved or being mixed in this type of mixture is large enough that can be seen by the naked eye without the aid of any device. A suspension mixture has a heterogeneous mixture.

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During physical activity of high intensity, your heart will beat at a percentage of its maximum rate. Whi

Maurinko [17]

Answer:

Explanation:

If it is of very high intensity it will be 85-100%

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3 years ago

g A lighter ball A with momentum PA=5.0 kg∙m/s in positive x-direction approaches a heavier ball B at rest before the collision.

Svet_ta [14]

Answer:

7kgm/s

Explanation:

Using the law of conservation of momentum which states that the sum of momentum of bodies before collision is equal to the sum of the bodies after collision.

Let P1A and P1B be the initial momentum of the bodies A and B respectively

Let P2A and P2B be the final momentum of the bodies A and B respectively after collision.

Based on the law:

P1A+P2A = P1B + P2B

Given P1A = 5kgm/s

P2A = 0kgm/s(ball B at rest before collision)

P2A = -2.0kgm/s (negative because it moves in the negative x direction)

P2B = ?

Substituting the values in the equation gives;

5+0 = -2+P2B

5+2 = P2B

P2B = 7kgm/s

3 0

3 years ago

A positive kaon (K+) has a rest mass of 494 MeV/c² , whereas a proton has a rest mass of 938 MeV/c². If a kaon has a total energ

vitfil [10]

Answer:

Explanation:

Rest mass of Kaon $M_{0K}$ = 494 MeV/c²

Rest mass of proton $M_{0P}$ = 938 MeV/c²

The rest energy is gotten by multiplying the rest mass by the square of the speed of light c²

for the kaon, rest energy $E_{0K}$ = 494c² MeV

for the proton, rest energy $E_{0P}$ = 938c² MeV

Recall that the rest energy, and the total energy are related by..

$E$ = γ $E_{0}$

which can be written in this case as

$E_{K}$ = γ $E_{0K}$ ...... equ 1

where $E$ = total energy of the kaon, and

$E_{0}$ = rest energy of the kaon

γ = relativistic factor = $\frac{1}{\sqrt{1 - \beta ^{2} } }$

where $\beta = \frac{v}{c}$

But, it is stated that the total energy of the kaon is equal to the rest mass of the proton or its equivalent rest energy, therefore...

$E_{K}$ = $E_{0P}$ ......equ 2

where $E_{K}$ is the total energy of the kaon, and

$E_{0P}$ is the rest energy of the proton.

From $E_{K}$ = $E_{0P}$ = 938c²

equ 1 becomes

938c² = γ494c²

γ = 938c²/494c² = 1.89

γ = $\frac{1}{\sqrt{1 - \beta ^{2} } }$ = 1.89

1.89 $\sqrt{1 - \beta ^{2} }$ = 1

squaring both sides, we get

3.57( 1 - $\beta^{2}$ ) = 1

3.57 - 3.57 $\beta^{2}$ = 1

2.57 = 3.57 $\beta^{2}$

$\beta^{2}$ = 2.57/3.57 = 0.72

$\beta = \sqrt{0.72}$ = 0.85

but, $\beta = \frac{v}{c}$

v/c = 0.85

v = <em>0.85c </em>

7 0

3 years ago

Racing greyhounds are capable of rounding corners at very high speeds. A typical greyhound track has turns that are 45 m diamete

Readme [11.4K]

Answer:

In m/s^2:

a=11.3778 m/s^2

In units of g:

a=1.161 g

Explanation:

Since the racing greyhounds are capable of rounding corners at very high speed so we are going use the following formula of acceleration for circular paths.

$a=\frac{v^2}{r}$