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Vadim26 [7]
3 years ago
7

1. Determine the magnitude of two equal but opposite charges if they attract one another with a force of 0.7N when at distance o

f 0.3m apart in a vacuum.​
Physics
1 answer:
adell [148]3 years ago
7 0

Answer:

q = 2.65 10⁻⁶ C

Explanation:

For this exercise we use Coulomb's law

        F =k \frac{q_1q_2}{r^2}

In this case they indicate that the load is of equal magnitude

       q₁ = q₂ = q

the force is attractive because the signs of the charges are opposite

       F = k \ \frac{q^2}{r^2}

       q = \sqrt{\frac{F \ r^2}{k} }

we calculate

        q = \sqrt{\frac{0.7 \ 0.3^2 }{9 \ 10^9}  }

        q = \sqrt{7 \ 10^{-12} }Ra 7 10-12

        q = 2.65 10⁻⁶ C

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Instead of moving back and forth, a conical pendulum moves in a circle at constant speed as its string traces out a cone (see fi
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Answer:

a

The  radial acceleration is  a_c  = 0.9574 m/s^2

b

The horizontal Tension is  T_x  = 0.3294 i  \ N

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Explanation:

The diagram illustrating this is shown on the first uploaded

From the question we are told that

   The length of the string is  L =  10.7 \ cm  =  0.107 \ m

     The mass of the bob is  m = 0.344 \  kg

     The angle made  by the string is  \theta  =  5.58^o

The centripetal force acting on the bob is mathematically represented as

         F  =  \frac{mv^2}{r}

Now From the diagram we see that this force is equivalent to

     F  =  Tsin \theta where T is the tension on the rope  and v is the linear velocity  

     So

          Tsin \theta  =   \frac{mv^2}{r}

Now the downward normal force acting on the bob is  mathematically represented as

          Tcos \theta = mg

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       \frac{Tsin \ttheta }{Tcos \theta }  =  \frac{\frac{mv^2}{r} }{mg}

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=>   g tan \theta  = \frac{v^2}{r}

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      a_c  =  \frac{v^2}{r}

=>  a_c  = gtan \theta

substituting values

     a_c  =  9.8  *  tan (5.58)

     a_c  = 0.9574 m/s^2

The horizontal component is mathematically represented as

     T_x  = Tsin \theta = ma_c

substituting value

   T_x  = 0.344 *  0.9574

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substituting value

     T_ y  =  0.344 * 9.8

      T_ y  = 3.2712 \ N

The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is  

         

       T  = T_x i  + T_y  j

substituting value  

      T  = [(0.3294) i  + (3.3712)j ] \  N

         

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Answer:

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U- 10 m/s

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S - 0.25 m

t -?

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