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3 years ago

10

Your science teacher gives you three liquids to pour into a jar. After pouring all of them into the jar, the liquids layer as se

en in the image. What property of matter prevents the three liquids from mixing together? answers: density texture color mass

2 answers:

viva [34]3 years ago

5 0

It is density because they are all not the same density

ANTONII [103]3 years ago

3 0

Answer:

Density

Explanation:

Each of the three liquids have different densities. Have you ever tried to mix oil and water? They don't mix because they have different densities.

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Four people return home from work and walk up the stairs to their own apartments in the same building which person is getting th

dmitriy555 [2]

Answer:

The person going to the highest apartment door.

Explanation:

Since gravitational potential energy, U = mgh where h is the height above the ground. The person with the highest gravitational potential energy would be the person going to the highest apartment door if we assume that their masses are the same.

6 0

4 years ago

Ask a testable question about each event listed in part a

Ratling [72]

Answer:

What?-

Explanation:

Where’s the story or whatever like that?

3 0

3 years ago

Item 8 The cost (in dollars) of making b bracelets is represented by 4 5b. The cost (in dollars) of making b necklaces is repres

grin007 [14]

Answer:

3b+2

Explanation:

First your question has so many mistakes

What I understand

bracelets cost 4+5b

necklaces cost 8b+6

First we have to make an equaion

=(8b+6)-(4+5b)

simplify it

=8b+6-4-5b

=3b+2 Answer

4 0

3 years ago

Hoop rolling up an inclined plane A hollow cylinder (or hoop) is rolling along a horizontal surface with speed v = 3.3 m/s when

Naya [18.7K]

Answer:

Explanation:

For rolling up or down an incline plane , the acceleration or deceleration of the rolling body is given by the following expression

a = g sinθ / (1 + k²/r² )

where k is radius of gyration of rolling body and θ is angle of inclination

a = g sin15 / ( 1 + 1 ) [ for hoop k = r ]

a = 9.8 x .2588 / 2

= 1.268 m / s²

a )

Let s be the distance up to which it goes

v² = u² - 2as

0 = 3.3² - 2 x 1.268 s

s = 4.3 m

b ) Let time in going up be t₁

v = u - at₁

0 = 3.3 - 1.268 t₁

t₁ = 2.6 s

Time in going down t₂

s = 1/2 a t₂²

4.3 = .5 x 1.268 t₂²

t₂ = 2.60

Total time

= t₁ +t₂

= 2.6 + 2.6

= 5.2 s

4 0

4 years ago

. (Use equations not the psychrometric chart) The dry- and wet-bulb temperatures of atmospheric air at 95 kPa are 25 and 17oC, r

Fantom [35]

Answer:

a) The specific humidity of air is $9.774\times 10^{-3}\,\frac{kg\,H_{2}O}{kg\,DA}$ .

b) The specific humidity of air is 0.464.

c) The dew-point temperature is 12.665 ºC.

Explanation:

a) The temperature of atmospheric air is considered the dry-bulb temperature, whereas the temperature of entirely saturated air is the the wet-bulb temperature. Dry bulb pressure is the atmospheric air. First we need to find the specific humidity at wet bulb temperature ( $\omega_{wb}$ ), measured in kilograms of water per kilogram of dry air:

$\omega_{wb} = \frac{0.622\cdot P_{wb}}{P_{db}-P_{wb}}$ (Eq. 1)

Where:

$P_{wb}$ - Wet bulb pressure, measured in kilopascals.

$P_{db}$ - Dry bulb pressure, measured in kilopascals.

Wet bulb pressure is the saturation pressure of water evaluated at wet bulb temperature, while dry bulb pressure in the pressure presented on statement. If $P_{db} = 95\,kPa$ and $P_{wb} = 1.9591\,kPa$ , then the specific humidity at wet bulb temperature is:

$\omega_{wb} = \frac{0.622\cdot (1.9591\,kPa)}{95\,kPa-1.9591\,kPa}$

$\omega_{wb} = 0.0131\,\frac{kg\,H_{2}O}{kg\,DA}$

Now we use the following equation to determine the dry bulb specific humidity ( $\omega_{db}$ ), measured in kilograms of water per kilogram of dry air:

$\omega_{db} = \frac{c_{p,a}\cdot (T_{wb}-T_{db})+\omega_{wb}\cdot h_{fg,wb}}{h_{g,db}-h_{f,wb}}$ (Eq. 2)

Where:

$c_{p,a}$ - Isobaric specific heat of air, measured in kilojoules per kilogram-Celsius.

$T_{wb}$ - Wet-bulb temperature, measured in Celsius.

$T_{db}$ - Dry-bulb temperature, measured in Celsius.

$\omega_{wb}$ - Wet-bulb specific humidity, measured in kilograms of water per kilogram of dry air.

$h_{fg,wb}$ - Wet-bulb specific enthalpy of vaporization of water, measured in kilojoules per kilogram.

$h_{g,db}$ - Dry-bulb specific enthalpy of saturated vapor, measured in kilojoules per kilogram.

$h_{f,wb}$ - Wet-bulb specific enthalpy of liquid vapor, measured in kilojoules per kilogram.

If we know that $T_{wb} = 17\,^{\circ}C$ , $T_{db} = 25\,^{\circ}C$ , $c_{p,a} = 1.005\,\frac{kJ}{kg\cdot ^{\circ}C}$ , $\omega_{wb} = 0.0131\,\frac{kg\,H_{2}O}{kg\,DA}$ , $h_{fg, wb} = 2460.6\,\frac{kJ}{kg}$ , $h_{g,db} = 2546.5\,\frac{kJ}{kg}$ and $h_{f,wb} = 71.355\,\frac{kJ}{kg}$ , the dry bulb specific humidity is:

$\omega_{db} = \frac{\left(1.005\,\frac{kJ}{kg\cdot ^{\circ}C} \right)\cdot (17\,^{\circ}C-25\,^{\circ}C)+\left(0.0131\,\frac{kg\,H_{2}O}{kg\,DA} \right)\cdot \left(2460.6\,\frac{kJ}{kg} \right)}{2546.5\,\frac{kJ}{kg}-71.355\,\frac{kJ}{kg} }$

$\omega_{db} = 9.774\times 10^{-3}\,\frac{kg\,H_{2}O}{kg\,DA}$

The specific humidity of air is $9.774\times 10^{-3}\,\frac{kg\,H_{2}O}{kg\,DA}$ .

b) Then, the relative humidity of air ( $\phi_{db}$ ), dimensionless, is obtained from this expression:

$\phi_{db} = \frac{\omega_{db}\cdot P_{db}}{(0.622+\omega_{db})\cdot P_{sat, db}}$ (Eq. 3)

Where $P_{sat, db}$ is the saturation pressure at dry-bulb temperature, measured in kilopascals.

If we know that $\omega_{db} = 9.774\times 10^{-3}\,\frac{kg\,H_{2}O}{kg\,DA}$ , $P_{db} = 95\,kPa$ and $P_{sat, db} = 3.1698\,kPa$ , the relative humidity of air is:

$\phi_{db} = \frac{\left(9.774\times 10^{-3}\,\frac{kg\,H_{2}O}{kg\,DA} \right)\cdot (95\,kPa)}{\left(0.622+9.774\times 10^{-3}\,\frac{kg\,H_{2}O}{kg\,DA}\right)\cdot 3.1698\,kPa}$

$\phi_{db} = 0.464$

The specific humidity of air is 0.464.

c) The dew point temperature is the temperature at which water is condensated when air is cooled at constant pressure. That temperature is equivalent to the saturation temperature at vapor pressure ( $P_{v}$ ), measured in kilopascals:

$P_{v} = \phi_{db} \cdot P_{sat, db}$ (Eq. 4)

( $\phi_{db} = 0.464$ , $P_{sat, db} = 3.1698\,kPa$ )

$P_{v} = 0.464\cdot (3.1698\,kPa)$

$P_{v} = 1.4707\,kPa$

The saturation temperature at given vapor pressure is:

$T_{dp} = 12.665\,^{\circ}C$

The dew-point temperature is 12.665 ºC.

4 0

4 years ago