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Leona [35]
4 years ago
5

In which wavelength region(s) would we need to carry out observations in order to study the accretion disk around a white dwarf

in a binary system?
Physics
1 answer:
Inga [223]4 years ago
7 0

In “Ultraviolet light” and “X-ray” light wavelength regions are needed to carry out observations to study the “accretion disk” around “a white dwarf” in a “binary system”.

<u>Explanation</u>:

Occurrence of an accretion disk is due to presence of diffuse material around a white dwarf in orbital motion. White dwarf is constituted from electron-degenerate matter therefore also called as degenerate dwarf in binary system. Frictional forces and gravitational forces are responsible to compress and high “temperature” of the material, results with emission of “electromagnetic radiation”. “Accretion disks” of white dwarf radiate in the UV-rays and X-rays wavelength region of spectrum.

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An electron moving parallel to a uniform electric field increases its speed from 2.0 × 107 m/s to 4.0 × 107 m/s over a distance
jeka94

Answer:

1.8\times 105 N/C

Explanation:

We are given that

u=2\times 10^7 m/s

v=4\times 10^7 m/s

d=1.9 cm=\frac{1.9}{100}=0.019 m

Using 1m=100 cm

We have to find the electric field strength.

v^2-u^2=2as

Using the formula

(4\times 10^7)^2-(2\times 10^7)^2=2a(0.019)

16\times 10^{14}-4\times 10^{14}=0.038a

0.038a=12\times 10^{14}

a=\frac{12}{0.038}\times 10^{14}=3.16\times 10^{16}m/s^2

q=1.6\times 10^{-19} C

Mass of electron,m=9.1\times 10^{-31} kg

E=\frac{ma}{q}

Substitute the values

E=\frac{9.1\times 10^{-31}\times 3.16\times 10^{16}}{1.6\times 10^{-19}}

E=1.8\times 105 N/C

7 0
4 years ago
True or false, the older rocks tend to be closer to the tear in the ocean floor
Natali5045456 [20]

Answer:

false!!

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4 0
3 years ago
Please help, I don't get this
Trava [24]
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5 0
3 years ago
A spring with a spring constant of 50 N/m is stretched 15cm. What is the force and energy associated with this stretching?
Olenka [21]
Data:
F (force) = ? (Newton)
k (<span>Constant spring force) = 50 N/m
x (</span>Spring deformation) = 15 cm → 0.15 m

Formula:
F = k*x

Solving: 
F = k*x
F = 50*0.15
\boxed{\boxed{F = 7.5\:N}}\end{array}}\qquad\quad\checkmark

Data:
E (energy) = ? (joule)
k (Constant spring force) = 50 N/m
x (Spring deformation) = 15 cm → 0.15 m

Formula:
E = \frac{k*x^2}{2}

Solving:(Energy associated with this stretching)
E = \frac{k*x^2}{2}
E =  \frac{50*0.15^2}{2}
E =  \frac{50*0.0225}{2}
E =  \frac{1.125}{2}
\boxed{\boxed{E = 0.5625\:J}}\end{array}}\qquad\quad\checkmark

7 0
3 years ago
Why do noble gasses rarely react with other elements?
ollegr [7]
Noble gasses have an outer shell full of electrons. A full outer energy level is the most stable arrangement of electrons. As a result, noble gases cannot become more stable by reacting with other elements and gaining or losing valence electrons. Therefore, noble gases are rarely involved in chemical reactions and almost never form compounds with other elements.
3 0
2 years ago
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