WILL GIVE BRAINLIEST! QUICK PLEASE HELP!!!

infrared radiation from young stars can pass through the heavy dust clouds surrounding them, allowing astronomers here on earth

Mashcka [7]

The wavelength of the infrared radiation is λ = $3.174$ × $10^{-5}$ m.

<h3>What is infrared radiation?</h3>

An infrared telescope is tuned to detect infrared radiation with a frequency of 9.45 THz.

We know that,

1 THz = 10¹² Hz

So,

f = 9.45 × 10¹² Hz

We need to find the wavelength of the infrared radiation.

λ=c/f

λ = 3× $10^{8}$ /9.45× $10^{12}$

λ = 3.174 × $10^{-5}$ m

The term "infrared radiation" (IR) refers to a part of the electromagnetic radiation spectrum with wavelengths between about 700 nanometers (nm) and one millimeter (mm). Longer than visible light waves but shorter than radio waves are infrared waves.

Electromagnetic radiation with wavelengths longer than those of visible light is known as infrared, also known as infrared light. Since it is undetectable to the human eye, The typical range of wavelengths considered to be infrared (IR) is from about 1 millimeter to the nominal red edge of the visible spectrum, or about 700 nanometers.

To learn more about infrared radiation from the given link:

brainly.com/question/13163856

#SPJ4

5 0

1 year ago

What is the correct order of the layers' density from lowest density to highest?

Orlov [11]

Answer:

C. crust, mantle, core

Explanation:

density increases as you travel from the crust to the inner core

the crust is on top

next is the mantle

and then the core

6 0

3 years ago

Read 2 more answers

Through what potential difference would you need to accelerate an alpha particle, starting from rest, so that it will just reach

Svetlanka [38]

Answer:

$\Delta V = 1.8 \times 10^7 V$

Explanation:

GIVEN

diameter = 15 fm = $15 \times 10^{-15}$ m

we use here energy conservation

$K_{i}+U_{i} =K_{f}+U_{f}$

there will be some initial kinetic energy but after collision kinetic energy will zero

$K_{i} + 0 = 0 + \frac{1}{4 \pi \epsilon _{0}} \frac{(2e)(92e)}{7.5 \times 10^{-15}}$

on solving these equations we get kinetic energy initial

$KE_{i} = 5.65\times 10 ^{-12} \times \frac {1 eV}{1.6 \times 10^{-19}}$

$KE_{i} = 35.33$ J ..............(i)

That is, the alpha particle must be fired with 35.33 MeV of kinetic energy. An alpha particle with charge q = 2 e

and gains kinetic energy K =e∆V ..........(ii)

by accelerating through a potential difference ∆V

Thus the alpha particle will

just reach the ${238}_U$ nucleus after being accelerated through a potential difference ∆V

equating (i) and second equation we get

e∆V = 35.33 Me V

$\Delta V = \frac{35.33}{2} MV\\\Delta V = 1.8 \times 10^7 V$

7 0

3 years ago

Consider a well-insulated rigid container with two chambers separated by a membrane. The total volume is 5.0 cubic meters. The f

mamaluj [8]

Answer:

The Entropy generated by the steam = 2.821 kJ/K

Explanation:

Total volume of container = 5m³

Heat transfer does not exist between system and surrounding, dQ = 0

At the first chamber, temperature of water at saturated liquid is 300°C

From the steam table:

Specific enthalpy of saturated liquid at 300°C , $h_{f} = 1344.8 kJ/kg$

Specific internal energy of saturated liquid at 300°C, $U_{f1} =$ 1332.7 kJ/kg

For closed system, the first law of thermodynamics state that:

dQ = dw + dU..................(1)

work done for free expansion, dw =0

0 = 0 + dU

dU = 0 , i.e. U₁ = U₂

At the second chamber,

The final pressure, P₂ = 50 kPa

From the steam table, at P₂ = 50 kPa, $U_{f2}$ = 340.49 kJ/kg

$(U_{fg} )_{2} = 2142.7 kJ/kg$

Let the dryness fraction at the second chamber = x

$U_{2} = U_{f2} + U_{fg2}$

$U_{2} = 340.49 + x2140.7$ Since U₁ = U₂

1332.7 = 340.49 + x2140.7

Dryness fraction, x = 0.463

From steam table, the specific volume is, $u_{f2} = 0.00103 m^{3} /kg\\$

$u_{2} = u_{f2} + xu_{fg2}$

$u_{2} = 0.00103 + 0.463(3.2393)\\u_{2} = 1.5 m^{3} /kg\\$

$u_{2} = \frac{v_{2} }{m_{2} }$

V₂ = 5 m³

1.5 = 5/m₂

m₂ = 3.33 kg

At 300°C $S_{1} = S_{f} = 3.2548 kJ/kg-k\\$

$S_{2} = S_{f2} + xS_{fg2}$

From the steam table,

$S_{f2} = 1.0912 kJ/kg-k\\S_{fg2} = 6.5019 kJ/kg-k\\S_{2} = 1.0912 + 0.463(6.5019)\\S_{2} = 4.102 kJ/kg-k$

Therefore the entropy generated will be :

Entropy = mass* (S₂ - S₁)

Entropy = 3.33* (4.102 - 3.2548)

Entropy = 2.821 kJ/K

5 0

3 years ago

Read 2 more answers

~~~!Here's the question!~~~

gtnhenbr [62]

The answer would most likely be A since obviously gravity weighs things down which helps the every other masses stay settled in place

6 0

3 years ago

Read 2 more answers