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3 years ago

J. J. Thomson s experiments provided evidence that an atom

1 answer:

5 0

J.J. Thomson's experiments provided evidence that an atom B.contains negatively charged particles.

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Often called velocity this is the velocity of an object at a particular moment in time

Yuliya22 [10]

Instantaneous speed?

8 0

3 years ago

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

3 years ago

Which of the following statements best describes chromatic aberration?

Rzqust [24]

Answer:

The answer is A/B, they're the same answer anyways.

Explanation:

Chromatic aberration is the result when the lens fail to focus all the colors on the same point. The light then focuses in different points,and could lead to causing two images at once. The main culprit of this is usually dispersion.

6 0

3 years ago

A solid non-conducting sphere of radius R carries a charge Q distributed uniformly throughout its volume. At a radius r (r <

Svet_ta [14]

Answer:

Hence the answer is E inside $= KQr_{1} /R^{3}$ .

Explanation:

E inside $= KQr_{1} /R^{3}$

so if r1 will be the same then

E $\begin{bmatrix}Blank Equation\end{bmatrix}$ proportional to 1/R3

so if R become 2R

E becomes 1/8 of the initial electric field.

8 0

2 years ago

Read 2 more answers

A seagull flies at a velocity of 9.00 m/s straight into the wind.

RideAnS [48]

a)If it takes the bird 18.0 minutes to fly 6 km away from the earth, the wind's speed will be 4 m/s.

b) The bird would need 7 minutes and 42 seconds to fly back 6 kilometers if he turned around and flew with the wind.

c)Compared to the 133.33 seconds it would take without the wind, the overall round-trip time is affected by the wind.

<h3>What is velocity?</h3>

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is

A seagull flies at a velocity, $\rm v_{SA} = 9 \ m/sec$

The time the bird takes,t=18.0 min

The distance traveled relative to the earth = 6.00 km

The seagull's relative velocity with reference to the ground as;

$\rm v_{sg} = \frac{6.00 \times 10^3 \ m }{(20 min) \times \frac{60 s }{1 \ min}} \\\\ v_{sg}= 5.00 \ m/sec$

Air velocity with reference to the ground is;

$\rm v_{AG}= v_{SG}-v_{SA} \\\\ v_{AG} = 5.00 \ m/sec - 9.00 \ m/sec \\\\ v_{AG} = -4.00 \ m/sec$

If the bird turns around and flies with the wind, The time will he take to return 6.00 km is;

$\rm v_{SG}=v_{SA}+v_{AG} \\\\ v_{SG}=-900 \ m/sec +(-4.00 \ m/sec) \\\\ v_{SG}= -13.00 \ m/sec$

The time the bird takes;

$\rm t = \frac{x_{SG}}{v_{SG}} \\\\ t = \frac{6.00 \times 10^3 \ m }{13.00 \ m/sec } \\\\ t = 462 m/sec \\\\ t = 7 \ min \ and \ 42 \ sec$

c)\

The total round-trip time compared to what it would be with no wind. is;

$\rm t = 20 \ min( \frac{60 \ sec }{1 \ min} )+ 462 \ sec \\\\ t = 1200 \ sec +6 462 \ ec \\\\ t= 1662 \ sec$

The time for the round trip is;

$\rm t = \frac{12 \times 10^ 3 }{ 9 \ m/sec } \\\\ t = 1333.33 \ sec$

Hence the wind's speed, the time bird would need to fly back the total round-trip time will be 4 m/s, 7 minutes and 42 seconds and 1333.33 sec.

To learn more about the velocity, refer to the link: brainly.com/question/862972.

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4 0

2 years ago