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Bess [88]
3 years ago
13

3. A -4.00-uC charge lies 20.0cm to the right of a 2.00-uC charge on the x axis. What is the force on the 2.00-uC charge?

Physics
1 answer:
Sladkaya [172]3 years ago
3 0

<u>Answer:</u>

The force on the 2.00-uC charge is 36 \times 10^{10} \mathrm{N}

<u>Explanation:</u>

We know that force between two charges is given by Coulomb’s law,

\mathrm{F}=\mathrm{k} \frac{q 1 q 2}{r * r}

Where k = Coulomb’s constant =

9.0 \times 10^{9} \mathrm{Nm}^{2} \mathrm{C}^{-2}

And q1 and q2 are the charges given to be = -4.00-uC and 2.00-uC charges

And r = distance between the charges = 20 cm = 0.2 m  

Substituting the given values in the formula we get force applied on 2.00\ \mu C charge,

F =  36 \times 10^{10} \mathrm{N} attractive force which is the required answer.

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An object is dropped from a height of 75.0 m above ground level. (a) Determine the distance traveled during the first second. (b
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Answer:

a)Distance traveled during the first second = 4.905 m.

b)Final velocity at which the object hits the ground = 38.36 m/s

c)Distance traveled during the last second of motion before hitting the ground = 33.45 m

Explanation:

a) We have equation of motion

             S = ut + 0.5at²

     Here u = 0, and a = g

              S = 0.5gt²

    Distance traveled during the first second ( t =1 )

              S = 0.5 x 9.81 x 1² = 4.905 m

   Distance traveled during the first second = 4.905 m.

b)  We have equation of motion

            v² = u² + 2as

      Here u = 0, s= 75 m and a = g

           v² = 0² + 2 x g x 75 = 150 x 9.81

           v = 38.36 m/s

      Final velocity at which the object hits the ground = 38.36 m/s

c) We have S = 0.5gt²

                   75 = 0.5 x 9.81 x t²

                    t = 3.91 s

   We need to find distance traveled last second

   That is

          S = 0.5 x 9.81 x 3.91² - 0.5 x 9.81 x 2.91² = 33.45 m

   Distance traveled during the last second of motion before hitting the ground = 33.45 m

       

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Answer:

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