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astraxan [27]
3 years ago
12

A student squeezes several oranges to make a glass of orange juice. The juice contains pieces of orange pulp mixed with the juic

e. Explain why this drink can be considered a combination of a suspension and a solution.
Physics
2 answers:
Mars2501 [29]3 years ago
7 0

<span>In </span>chemistry<span>, a suspension<span> is a heterogeneous mixture containing solid particles that are sufficiently large for </span></span>sedimentation<span>. Usually they must be larger than one </span>micrometer<span>. A Suspension is a heterogeneous mixture in which the solute particles do not dissolve but get suspended throughout the bulk of the medium. a </span>solution<span> is a </span>homogeneous mixture<span> <span>composed of two or more substances. In such a mixture, a </span>solute<span> is a substance </span></span>dissolved<span> <span>in another substance, known as a </span>solvent.. it is considered a solution and suspension because the orange juice mix with water while the pulp did not and retained as solid</span>

nasty-shy [4]3 years ago
7 0

Answer:

The juice contains sugars, plant pigments, and other chemicals dissolved in water. This is a solution.  The pieces of orange pulp will rise to the top or settle to the bottom of the juice if it is allowed to sit.  The pieces of pulp mixed with the juice form a suspension.

Explanation:

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red has a longer wavelength than yellow. Yellow has a longer wavelength than green.

Explanation

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Hurricanes are especially destructive because of the abrupt decrease in air pressure that they bring with the winds. Consider a
ivann1987 [24]

Answer:

2.96 × 10^4 N

Explanation:

1 atm = 101325 N/m², pressure inside the airtight room = 1.02 atm, pressure outside due to hurricane = 0.91 atm

net pressure directed outward = P inside - P  outside

net pressure = 1.02 - 0.91 = 0.11 atm

where 1 atm = 101325N/m²

0.11 atm = 0.11 × 101325 N/m² = 11145.75 N/m²

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8 0
3 years ago
At a rock concert, the sound intensity 1.0 m in front of the bank of loudspeakers is 0.10 W/m². A fan is 30 m from the loudspeak
Klio2033 [76]

To solve this problem we will apply the concepts related to the Area, the power and the proportionality relationships between intensity and distance.

The expression for sound power is,

P = AI

Here,

A = Area

I = Intensity

P = Power

At the same time the area can be written as,

A = \frac{\pi d^2}{4}

Now the intensity is inversely proportional to the square of the distance from the source, then

I \propto \frac{1}{r^2}

The expression for the intensity at different distance is

\frac{I_1}{I_2}= \frac{r^2_2}{r_1^2}

Here,

I_1 = Intensity at distance 1

I_2 = Intensity at distance 2

r_1 = Distance 1 from light source

r_2 = Distance 2 from the light source

If we rearrange the expression to find the intensity at second position we have,

I_2 = I_1 (\frac{r_1^2}{r_2^2})

If we replace with our values at this equation we have,

I_2 = (0.10W/m^2)(\frac{1.0m^2}{30.0m^2})

I_2 = 1.11*10^{-4} W/m^2

Now using the equation to find the area we have that

A = \frac{\pi (8.4*10^{-3}m)^2}{4}

A = 5.5*10^{-5}m^2

Finally with the intensity and the area we can find the sound power, which is

P = AI

P = (5.5*10^{-5}m^2)(1.11*10^{-4}W/m^2)

P = 6.1*10^{-9}J/s

Power is defined as the quantity of Energy per second, then

E = 6.1*10^{-9}J

8 0
3 years ago
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