Which scenario requires the most power?

Match the words in the left-hand column to the appropriate blank in the sentences in the right-hand column. Use each word only o

Vlad [161]

Answer:

1. About a trillion comets are thought to be located far, far beyond Pluto in the Oort cloud.

2. The bright spherical part of a comet observed when it is close to the Sun is the coma .

3. A comet's plasma tail stretches directly away from the Sun.

4. A comet's nucleus is the frozen portion of a comet.

5. Particles ejected from a comet can cause a(n) meteor shower on Earth.

6. The Kuiper belt extends from about beyond the orbit of Neptune to about twice the distance of Neptune from the Sun.

Explanation:

1. Oort cloud is a theoretical cloud of icy objects in interstellar space. Its inner limit begins about 2000-5000 AU from sun outer limit (about 100000-200000 AU) defines the boundary of solar system.

Note: AU is the astronomical unit which is defined as the distance between the Sun and Earth.

2. Comets are described as “dirty snowballs” which is a mixture of frozen gases and dust. When this snowball gets closer to the Sun, gas evaporates, and a fuzzy haze of gas and dust emerges. This fuzzy tail is the coma.

3. Dust and gas in the coma interact with the sunlight and the solar wind. Charged particles in the coma are caught up in the magnetic field of sun and flow away from the sun.

4. Nucleus is the dirty snowball which consists of rock, dust and frozen gas.

5. Meteor shower is a celestial event with number of meteors can be observed. Meteors are meteoroids that enters earth’s atmosphere with extremely high speeds.

6. Kuiper belt is a region beyond Neptune including dwarf planets and icy objects.

5 0

2 years ago

A vertical spring stretches 4.0 cm when a 12-g object is hung from it. The object is replaced with a block of mass 28 g that osc

marin [14]

Answer:

Time period of the osculation will be 0.0671 sec

Explanation:

It is given a vertical spring is stretched by 4 cm

So change in length of the spring x = 4 cm = 0.04 m

Mass which is hung from it m = 12 gram = 0.012 kg

Sprig force will be equal to weight of the mass

So $kx=mg$

$k\times 0.04=0.012\times 9.8$

k = 244.7 N/m

Now new mass is m = 28 gram = 0.028 kg

So time period with new mass will be

$T=2\pi \sqrt{\frac{m}{k}}$

$=2\times 3.14 \sqrt{\frac{0.028}{244.7}}=0.0671sec$

4 0

2 years ago

Somebody please help me with this question..!

Rudik [331]

8.1) Here, Pressure increases with depth in the ocean at higher rate then that of increase in altitude in atmosphere. So, the rate of change of pressure is different in these journeys.So, your most correct answer would be option D.

8.2) Difference in the shape of lines is due to different density of air & water. So, that physical property and your answer would be option A

So, in Short Answers of your questions are:
8.1) - Option D
8.2) - Option A

Hope this helps!

3 0

2 years ago

Which letter on the diagram below represents the trough of a wave?

Nikitich [7]

A would be the wavelength, C would be a crest, D would be the amplitude, leaving B which is the trough.

8 0

2 years ago

Read 2 more answers

Compute the expected shell-model quadrupole moment of 209Bi () and compare with the experimental value, - 0.37 b

Over [174]

Answer:

0.22 b

Explanation:

Quadrupole moment of the nucleon is,

$Q=-\frac{2j-1}{2(j+1)}\frac{3}{5}R^{2}$

And also,

$R^{2}=R^{2} _{0}A^{\frac{2}{3} }$

And, $R _{0}=1.2\times 10^{-15}m$

Now,

$Q=-\frac{2j-1}{2(j+1)}\frac{3}{5}R^{2} _{0}A^{\frac{2}{3} }$

For Bismuth $j=\frac{9}{2}$ and A is 209.

$Q=-\frac{2\frac{9}{2} -1}{2(\frac{9}{2} +1)}\frac{3}{5}(1.2\times 10^{-15}) ^{2}(209)^{\frac{2}{3} }\\Q=0.628\times 35.28\times 10^{-30} \\Q=22.15\times 10^{-30} m^{2} \\Q=0.2215\times 10^{-28} m^{2} \\Q=0.22 barn$

Therefore, the expected value of quadrupole is 0.22 b which is quite related with experimental value which is 0.37 b

3 0

2 years ago