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Jlenok [28]
3 years ago
15

You observe that a negatively charged plastic pen repels a charged piece of magic tape. You then observe that the same piece of

tape is repelled when brought near a small metal sphere. You are wearing rubber soled shoes, and you touch the metal sphere with your hand. After you touch the metal sphere, you observe that the tape is attracted to the metal sphere. Which of the following statements could be true. List all that apply. (1) Electrons from the sphere moved into the salt water on your skin, where they reacted with sodium ions. (2) The excess negative charge from the sphere spread out all over your body. (3) Chloride ions from the salt water on your hand moved onto the sphere. (4) Sodium ions from the salt water on your hand moved onto the sphere. (5) Electrons from your hand moved onto the sphere. (6) Electrons from the sphere traveled through your body into the Earth. (7) After you touched it, the metal sphere was very nearly neutral.
Physics
1 answer:
hichkok12 [17]3 years ago
5 0

Answer:

(2) The excess negative charge from the sphere spread out all over your body.

(7) After you touched it, the metal sphere was very nearly neutral.

Explanation:

Plastic pen repels magic tape so magic tape is also  negatively charged . Further , magic tape repels small metal sphere that means small sphere also is negatively charged.

Now when small sphere is touched by a man insulated from ground , the charge is distributed between man and small sphere according to their capacitance .

Since human body will have greater capacitance ,it will acquire larger share of charge . Sphere being of very small size will retain very less charge and it will become almost neutral . Hence it will be   attracted by  charged tape .

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Explanation:

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Compare the circular velocity of a particle orbiting in the Encke Division, whose distance from Saturn 133,370 km, to a particle
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Answer:

The particle in the D ring is 1399 times faster than the particle in the Encke Division.

Explanation:

The circular velocity is define as:

v = \frac{2 \pi r}{T}  

Where r is the radius of the trajectory and T is the orbital period

To determine the circular velocity of both particles it is necessary to know the orbital period of each one. That can be done by means of the Kepler’s third law:

T^{2} = r^{3}

Where T is orbital period and r is the radius of the trajectory.

Case for the particle in the Encke Division:

T^{2} = r^{3}

T = \sqrt{(133370 Km)^{3}}

T = \sqrt{(2.372x10^{15} Km)}

T = 4.870x10^{7} Km

It is necessary to pass from kilometers to astronomical unit (AU), where 1 AU is equivalent to 150.000.000 Km ( 1.50x10^{8} Km )

1 AU is defined as the distance between the earth and the sun.

\frac{4.870x10^{7} Km}{1.50x10^{8}Km} . 1AU

T = 0.324 AU

But 1 year is equivalent to 1 AU according with Kepler’s third law, since 1 year is the orbital period of the earth.

T = \frac{0.324 AU}{1 AU} . 1 year

T = 0.324 year

That can be expressed in units of days

T = \frac{0.324 year}{1 year} . 365.25 days  

T = 118.60 days

<em>Circular velocity for the particle in the </em><em>Encke Division</em><em>:</em>

v = \frac{2 \pi r}{T}

v = \frac{2 \pi (133370 Km)}{(118.60 days)}

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

118.60 days .\frac{86400 s}{1 day} ⇒ 10247040 s

133370 Km .\frac{1000 m}{1 Km} ⇒ 133370000 m

v = \frac{2 \pi (133370000 m)}{(10247040 s)}

v = 81.778 m/s

Case for the particle in the D Ring:

For the case of the particle in the D Ring, the same approach used above can be followed

T^{2} = r^{3}

T = \sqrt{(69000 Km)^{3}}

T = \sqrt{(3.285x10^{14} Km)}

T = 1.812x10^{7} Km

\frac{1.812x10^{7} Km}{1.50x10^{8}Km} . 1 AU

T = 0.120 AU

T = \frac{0.120 AU}{1 AU} . 1 year

T = 0.120 year

T = \frac{0.120 year}{1 year} . 365.25 days  

T = 43.83 days

<em>Circular velocity for the particle in </em><em>D Ring</em><em>:</em>

v = \frac{2 \pi r}{T}

v = \frac{2 \pi (69000 Km)}{(43.83 days)}

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

43.83 days . \frac{86400 s}{1 day} ⇒ 3786912 s

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v = \frac{2 \pi (69000000 m)}{(3786912 s)}

v = 114.483 m/s

 

\frac{114.483 m/s}{81.778 m/s} = 1.399            

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Tju [1.3M]

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Sherpas do work that is much more meaningful than the work other climbers do.

Explanation:

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tensional force

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