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4 years ago

Which list contains only objects that orbit the sun in our solar system?

Physics

1 answer:

Andre45 [30]4 years ago

3 0

The answer would be C.

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Two sound waves, from two different sources with the same frequency, 540 Hz, travel in the same direction at 330 m s . The sourc

oee [108]

Answer:

The value is $\Delta \phi = 4.12 \ rad$

Explanation:

From the question we are told that

The frequency of each sound is $f_1 = f_2 = f = 540 \ Hz$

The speed of the sounds is $v = 330 \ m/s$

The distance of the first source from the point considered is $a = 4.40 \ m$

The distance of the second source from the point considered is $b = 4.00 \ m$

Generally the phase angle made by the first sound wave at the considered point is mathematically represented as

$\phi_a = 2 \pi [\frac{a}{\lambda} + ft]$

Generally the phase angle made by the first sound wave at the considered point is mathematically represented as

$\phi_b = 2 \pi [\frac{b}{\lambda} + ft]$

Here b is the distance o f the first wave from the considered point

Gnerally the phase diffencence is mathematically represented as

$\Delta \phi= \phi_a - \phi_b = 2 \pi [\frac{ a}{\lambda} + ft ] - 2 \pi [\frac{b}{\lambda} + ft ]$

=> $\Delta \phi = \frac{2\pi [ a - b]}{ \lambda }$

Gnerally the wavelength is mathematically represented as

$\lambda = \frac{v}{f}$

=> $\lambda = \frac{330}{540}$

=> $\lambda = 0.611 \ m$

=> $\Delta \phi = \frac{2* 3.142 [ 4.40 - 4.0 ]}{ 0.611 }$

=> $\Delta \phi = 4.12 \ rad$

5 0

3 years ago

When atoms get more and more energy, they can ionize. What can happen to a gas as more energy is added to its atoms?

Ulleksa [173]

Answer: Option (A) is the correct answer.

Explanation:

When more and more energy is provided to a gas then its atoms move more rapidly.

This rapid and continuous movement converts the gas into hot ionized ions which have positively charged ions and negatively charged electrons.

Therefore, we can conclude that as the atoms move faster, the gas can change into a plasma.

5 0

3 years ago

Read 2 more answers

A graduated cylinder full of a clear liquid and small pieces of white material with white fumes emerging from it. Which indicato

Verdich [7]

Answer:

Explanation:

5 0

3 years ago

Read 2 more answers

What does a ‘0’ mean in binary code?

posledela

Answer:

<em>The answer is C, a switch is set so an electrical charge cannot flow.</em>

Explanation:

In a transistor (working as a switch), 0 represents no flow of electricity; 1 represents electricity being allowed to flow

6 0

3 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