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AVprozaik [17]
4 years ago
13

When you rub a plastic rod with fur, the plastic rod becomes negatively charged and the fur becomes positively charged. As a con

sequence of rubbing the rod with the fur,
A. the rod and fur both gain mass.
B. the rod and fur both lose mass.
C. the rod gains mass and the fur loses mass.
D. the rod loses mass and the fur gains mass.
E. none of the above
Physics
1 answer:
Yuri [45]4 years ago
4 0

Answer: C. the rod gains mass and the fur loses mass.

Explanation:Atomic particles have mass. The electron has a mass that is approximately 1/1836 that of the proton and with exchange exchange of charge this is also factored in. The movement of effect described above is known as the triboelectic charging process—charging by friction—which results in a transfer of electrons between the two objects when they are rubbed together. Plastic having a much greater affinity for electrons than animal fur pulls electrons from the atoms of fur, leaving both objects with an imbalance of charge. The plastic rod would have an excess of electrons and the fur has a shortage of electrons. Having an excess of electrons, the plastic is charged negatively and has more mass. In the same vein, the shortage of electrons on the fur leaves it with a positive charge and consequently with lesser mass.

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erma4kov [3.2K]

Answer:

The extent to which it would stretch  is \Delta L = 0.015 \ m

Explanation:

From the question we are told that

    The initial length is  L = 1.00m

     The area is  A = 0.500 mm^2 = \frac{0.500}{1 *10^6} = 0.500*10^6 \ m^2

     The Young modulus of the steel is  Y = 2.0*10^{11} Pa

     The tension   is  T =1500 N

The Young modulus is mathematically represented as

       Y = \frac{\sigma}{e}

Where \sigma is the stress which is mathematically represented as

           \sigma = \frac{F}{A}  

Substituting values

            \sigma = \frac{1500}{0.500*10^{-6}}  

           \sigma = 3.0*10^9 N/m^2  

And  e is the strain which is mathematically represented as

            e = \frac{\Delta L}{L }

Where \Delta L The extension of the steel string

Substituting these into the equation above

             Y = \frac{3.0*10^9}{\frac{\Delta L}{L} }

Substituting values  

           2.0 *10^{11} = \frac{3.0*10^9}{\frac{\Delta L}{L} }

          \Delta L = \frac{3.0*10^9  * 1}{2.0 *10^{11}}

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7 0
3 years ago
I NEED HELP PLEASE HELP ME
AnnZ [28]

Answer:

24000N

Explanation:

The mass of the object stays the same, regardless of its gravitational field strength. But, weight=mass*gravitational field strength.

So, the gravitational field strength on Earth is 10N. 2400*10=24000.

hope this helps

6 0
3 years ago
Energy does / does not change the phase of matter.
kati45 [8]

Answer:

does

Explanation:

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The magnetic circuit below is excited by a 100-turn coil wound over the central leg. The mean length of the central leg is 5.5cm
Dafna11 [192]

Answer:

Hello your question is incomplete attached below is the complete question and solution

<em>answer; </em>

attached below

Explanation:

<em>Given data:</em>

100-turn coil

mean length of central leg = 5.5 cm

mean length of outer paths = 15.5 cm

relative permeability = 2000

cross sectional area ( A ) = 1 cm^2

distance x = 1 cm  

attached below is a detailed solution

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3 years ago
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Answer:

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Explanation:

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