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3 years ago

Red-orange sunsets and blue skies are a result of what type of light-matter interaction?

Physics

2 answers:

fomenos3 years ago

8 0

B. polarization
i hope that is correct

kotykmax [81]3 years ago

4 0

Polarization is the correct answer.

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A rigid, insulated tank whose volume is 10 L is initially evacuated. A pinhole leak develops and air from the surroundings at 1

balandron [24]

Answer:

The answer is " $143.74^{\circ} \ C , 8.36\ g, and \ 2.77\ \frac{K}{J}$ "

Explanation:

For point a:

Energy balance equation:

$\frac{dU}{dt}= Q-Wm_ih_i-m_eh_e\\\\$

$W=0\\\\Q=0\\\\m_e=0$

From the above equation:

$\frac{dU}{dt}=0-0+m_ih_i-0\\\\\Delta U=\int^{2}_{1}m_ih_idt\\\\$

because the rate of air entering the tank that is $h_i$ constant.

$\Delta U = h_i \int^{2}_{1} m_i dt \\\\= h_i(m_2 -m_1)\\\\m_2u_2-m_1u_2=h_i(M_2-m_1)\\\\$

Since the tank was initially empty and the inlet is constant hence, $m_2u-0=h_1(m_2-0)\\\\m_2u_2=h_1m_2\\\\u_2=h_1\\\\$

Interpolate the enthalpy between $T = 300 \ K \ and\ T=295\ K$ . The surrounding air

temperature:

$T_1= 25^{\circ}\ C\ (298.15 \ K)\\\\\frac{h_{300 \ K}-h_{295\ K}}{300-295}= \frac{h_{300 \ K}-h_{1}}{300-295.15}$

Substituting the value from ideal gas:

$\frac{300.19-295.17}{300-295}=\frac{300.19-h_{i}}{300-298.15}\\\\h_i= 298.332 \ \frac{kJ}{kg}\\\\Now,\\\\h_i=u_2\\\\u_2=h_i=298.33\ \frac{kJ}{kg}$

Follow the ideal gas table.

The $u_2= 298.33\ \frac{kJ}{kg}$ and between temperature $T =410 \ K \ and\ T=240\ K.$

Interpolate

$\frac{420-410}{u_{240\ k} -u_{410\ k}}=\frac{420-T_2}{u_{420 k}-u_2}$

Substitute values from the table.

$\frac{420-410}{300.69-293.43}=\frac{420-T_2}{{u_{420 k}-u_2}}\\\\T_2=416.74\ K\\\\=143.74^{\circ} \ C\\\\$

For point b:

Consider the ideal gas equation. therefore, p is pressure, V is the volume, m is mass of gas. $\bar{R} \ is\ \frac{R}{M}$ (M is the molar mass of the gas that is $28.97 \ \frac{kg}{mol}$ and R is gas constant), and T is the temperature.

$n=\frac{pV}{TR}\\\\$

$=\frac{(1.01 \times 10^5 \ Pa) \times (10\ L) (\frac{10^{-3} \ m^3}{1\ L})}{(416.74 K) (\frac{8.314 \frac{J}{mol.k} }{2897\ \frac{kg}{mol})}}\\\\=8.36\ g\\\\$

For point c:

Entropy is given by the following formula:

$\Delta S = mC_v \In \frac{T_2}{T_1}\\\\=0.00836 \ kg \times 1.005 \times 10^{3} \In (\frac{416.74\ K}{298.15\ K})\\\\=2.77 \ \frac{J}{K}$

5 0

2 years ago

two forces x and y are acting at 120 degrees to each other, if the magnitude are 8N and 10N respectively determine their resulta

Ber [7]

The resultant force of both forces is 15.62 N.

<h3 /><h3>What is resultant?</h3>

The Resultant of forces is a single force obtained when two or more forces are combined.

To calculate the resultant of the force, we use the formula below.

Formula:

R = √[a²+b²-2abcos∅]..................... Equation 1

Where:

R = Resultant of the forces.
∅ = Angle between both forces

From the question,

Given:

a = 8 N
b = 10 N

Substitute these values into equation 1

R = √[8²+10²-2×8×10cos120°]
R = √[64+100-160cos120°]
R =√ [164-160(-0.5)]
R = √[164+80]
R = √(244)
R = 15.62 N

Hence, the resultant force of both forces is 15.62 N.

Learn more about resultant force here: brainly.com/question/25239010

#SPJ1

8 0

2 years ago

Read 2 more answers

Simon is writing a story about an astronaut whose spacecraft has been boarded by space pirates. The astronaut has her lucky penn

scZoUnD [109]

Oct 27, 2015 - In The Martian, astronaut Mark Watney and his crewmates onboard ... is grossly unqualified to challenge the science behind The Martian. ... of Space on the Law of the Sea, that Treaty has thus far been signed by only a handful of nations, none of which have their own ability to launch astronauts into space ...

4 0

3 years ago

Your low-flow showerhead is delivering water at 1.2×10−4m3/s, about 2.0 gallons per minute.

Oksanka [162]

To solve this problem it is necessary to apply the fluid mechanics equations related to continuity, for which the proportion of the input flow is equal to the output flow, in other words:

$Q_1 = Q_2$

We know that the flow rate is equivalent to the velocity of the fluid in its area, that is,

$Q = VA$

Where

V = Velocity

A = Cross-sectional Area

Our values are given as

$Q_2 = 1.2*10^{-4}m^3/s$

$d = 0.021m$

$r = \frac{0.021}{2} = 0.0105m$

Since there is continuity we have now that,

$V_1A_1 = Q_2$

$V_1A_1 = 1.2*10^{-4}$

$V_1 = \frac{1.2*10^{-4}}{A_2}$

$V_1 = \frac{1.2*10^{-4}}{\pi r^2}$

$V_1 = \frac{1.2*10^{-4}}{\pi (0.0105)^2}$

$V_1 =0.347m/s$

Therefore the speed of the water's house supply line is 0.347m/s

7 0

3 years ago

A rocket is launched straight up from the earth's surface at a speed of 1.60×104 m/s . what is its speed when it is very far awa

AlladinOne [14]

16,000 m/s
Since it’s speed, and the distance is unknown. Gravity isn’t applying a noticeable force too on the rocket, as if it were, then the rocket would be accelerating negatively.

7 0

3 years ago