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dlinn [17]
2 years ago
8

Please answer this question ​

Physics
1 answer:
steposvetlana [31]2 years ago
5 0

Hope you understand

Have a nice day

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Two masses are attracted by a gravitational force of 7 N.
damaskus [11]

Answer:

The answer to your question is: F  = 0.4375 N. The force will be 16 times lower than with the first conditions.

Explanation:

Data

F = 7 N

F = ?  if the masses is quartered

Formula

F = \frac{Km1m2}{r2}

Process

Normal conditions F = Km₁m₂/r²  = 7              

When masses quartered        F = K(m₁/4)(m₂/4)/r²  = ?

                                                F = K(m₁m₂/16)/r²

                                                F = K(m₁m₂/16r²      = 7/16  = 0.4375 N

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What is the strength of the electric field 0.1 mmmm below the center of the bottom surface of the plate
Aleksandr [31]

Complete Question

A thin, horizontal, 12-cm-diameter copper plate is charged to 4.4 nC . Assume that the electrons are uniformly distributed on the surface. What is the strength of the electric field 0.1 mm above the center of the top surface of the plate?

Answer:

The  values is  E =248.2 \  N/C

Explanation:

From the question we are told that

   The  diameter is  d =  12 \  cm  =  0.12 \ m

    The charge  is  Q =  4.4 nC  =  4.4 *10^{-9} \  C

    The  distance from the center  is  k =  0.1 \ mm   =  1*10^{-4} \  m

Generally the radius is mathematically represented as

        r =  \frac{d}{2}

=>     r =  \frac{0.12}{2}

=>       r =  0.06 \  m

Generally electric field is mathematically represented as  

       E =  \frac{Q}{ 2\epsilon_o } [1 - \frac{k}{\sqrt{r^2 +  k^2 } } ]

substituting values  

      E =  \frac{4.4 *10^{-9}}{ 2* (8.85*10^{-12}) } [1 - \frac{(1.00 *10^{-4})}{\sqrt{(0.06)^2 +  (1.0*10^{-4})^2 } } ]

     E =248.2 \  N/C

4 0
4 years ago
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