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3 years ago

Which wavelength of light is capable of penetrating the dust of a nebula?

1 answer:

ivann1987 [24]3 years ago

8 0

When the dust is too thick to penetrate with visible light, such as the Nebula, Radio Waves are used to penetrate the dust. Longer radio waves can completely penetrate the thick cloud cover, allowing scientists to beam radar waves.

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The brakes of a 125 kg sled are applied while it is moving at 8.1 m/s, which exerts a force of 261 N to slow the sled down. How

sammy [17]

Answer:

15.7 m

Explanation:

m = mass of the sled = 125 kg

v₀ = initial speed of the sled = 8.1 m/s

v = final speed of sled = 0 m/s

F = force applied by the brakes in opposite direction of motion = 261

d = stopping distance for the sled

Using work-change in kinetic energy theorem

- F d = (0.5) m (v² - v₀²)

- (261) d = (0.5) (125) (0² - 8.1²)

d = 15.7 m

6 0

2 years ago

X rays of wavelength 0.0169 nm are directed in the positive direction of an x axis onto a target containing loosely bound electr

mamaluj [8]

Answer:

a) 4.04*10^-12m

b) 0.0209nm

c) 0.253MeV

Explanation:

The formula for Compton's scattering is given by:

$\Delta \lambda=\lambda_f-\lambda_i=\frac{h}{m_oc}(1-cos\theta)$

where h is the Planck's constant, m is the mass of the electron and c is the speed of light.

a) by replacing in the formula you obtain the Compton shift:

$\Delta \lambda=\frac{6.62*10^{-34}Js}{(9.1*10^{-31}kg)(3*10^8m/s)}(1-cos132\°)=4.04*10^{-12}m$

b) The change in photon energy is given by:

$\Delta E=E_f-E_i=h\frac{c}{\lambda_f}-h\frac{c}{\lambda_i}=hc(\frac{1}{\lambda_f}-\frac{1}{\lambda_i})\\\\\lambda_f=4.04*10^{-12}m +\lambda_i=4.04*10^{-12}m+(0.0169*10^{-9}m)=2.09*10^{-11}m=0.0209nm$

c) The electron Compton wavelength is 2.43 × 10-12 m. Hence you can use the Broglie's relation to compute the momentum of the electron and then the kinetic energy.

$P=\frac{h}{\lambda_e}=\frac{6.62*10^{-34}Js}{2.43*10^{-12}m}=2.72*10^{-22}kgm\\$

$E_e=\frac{p^2}{2m_e}=\frac{(2.72*10^{-22}kgm)^2}{2(9.1*10^{-31}kg)}=4.06*10^{-14}J\\\\1J=6.242*10^{18}eV\\\\E_e=4.06*10^{-14}(6.242*10^{18}eV)=0.253MeV$

5 0

3 years ago

"In a pure substance all the particles must be identical; therefore pure substances are composed only of elements." Do you agree

kenny6666 [7]

I do not agree with the statement.
The "substance" can be a compound. It's "pure"
as long as there's nothing else in it but its name.

'Pure' water is 100% H₂O with nothing else in it.
'Pure' table salt is 100% NaCl with nothing else in it.
'Pure' carbon dioxide is 100% CO₂ with nothing else in it.

These example substances are all compounds, not elements.

6 0

2 years ago

Before you conduct your experiment, you need to form a hypothesis. A hypothesis is a prediction of what you think will happen in

Stolb23 [73]

Example: A apple rotting.

If I put my apple in a fridge, then it would not rot as fast because it is in a cooled area. (example)

Hope it helps! Brainiest Answer would be amazing!

5 0

3 years ago

NEED HELP NOW WILL GIVE BRAINLIEST

Arturiano [62]

Answer:

B. solar flares sunspots, solar flares prominences

Explanation:

Some features of the Sun's surface include sunspots, solar flares, and prominences

pls mark me brainliest its the right answer

6 0

2 years ago

Read 2 more answers