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LekaFEV [45]
1 year ago
5

Select the correct answer. What happens when a negatively charged object A is brought near a neutral object B? A. Object B gets

a negative charge. B. Object B gets a positive charge. C. Object B stays neutral but becomes polarized. D. Object A gets a positive charge. E. Object A loses all its charge.
Physics
1 answer:
givi [52]1 year ago
4 0

What happens when a negatively charged object A is brought near a neutral object B is that: B. Object B gets a positive charge.

<h3>The law of electrostatic forces.</h3>

According to the law of electrostatic forces, unlike charges attract each other while like charges repel one another.

This ultimately implies that, objects that are having the same charges (like charges) would repel one another, and this causes a transfer of electrons (charges) to any differently charged object which comes in contact with it, through a process known as conduction.

In this context, we can reasonably and logically deduce that the negative charge of object A would induce an opposite charge (positive) on object B when a negatively charged object A is brought near a neutral object B.

Read more on charges here: brainly.com/question/1824478

#SPJ1

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Answer:

The deceleration is  a =  - 76.27 m/s^2

Explanation:

From the question we are told that

   The height above  firefighter safety net is H  = 14 \ m

   The length by which the net is stretched is s =  1.8 \ m

   

From the law of energy conservation

    KE_T + PE_T =  KE_B + PE_B

 Where KE_T is the kinetic energy of the person before jumping which equal to zero(because to kinetic energy at maximum height )

   and  PE_T is the potential energy of the before jumping  which is mathematically represented at

          PE_T  = mg H

and  KE_B is the kinetic energy of the person just before landing on the safety net  which is mathematically represented at

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and  PE_B is the potential energy of the person as he lands on the safety net which has a value of zero (because it is converted to kinetic energy )

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          mgH =  \frac{1}{2} m v^2

=>           v =  \sqrt{2 gH }

    substituting values

                v =  16.57 m/s

Applying the equation o motion

             v_f =  v  + 2 a s

Now the final velocity is zero because the person comes to rest

      So

         0 = 16.57 + 2 * a * 1.8

            a =  - \frac{16.57^2 }{2 * 1.8}

            a =  - 76.27 m/s^2

         

         

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