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larisa [96]
2 years ago
6

A conducting sphere 45 cm in diameter carries an excess of charge, and no other charges are present. You measure the potential o

f the surface of this sphere and find it to be 14 kV relative to infinity. Find the excess charge on this sphere.
Physics
1 answer:
uysha [10]2 years ago
7 0

Answer:

The excess charge is Q  = 3.5 *10^{-7} \ C

Explanation:

From the question we are told that

      The diameter is d = 45 \ cm  =  0.45 \ m

       The potential of the surface is V =  14 \ kV   =  14 *10^{3} \ V

     

The radius of the sphere is  

           r = \frac{d}{2}

substituting values

          r = \frac{0.45}{2}

         r = 0.225 \ m

The potential on the surface is mathematically represented as  

          V  =  \frac{k  *  Q  }{r }

Where k is coulomb's constant with value  k  =  9*10^{9}  \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.

  given from the question that there is no other charge the Q is the excess charge  

Thus  

        Q  =  \frac{V* r}{ k}

substituting values

        Q  =  \frac{14 *10^{3}  0.225}{ 9*10^9}

        Q  = 3.5 *10^{-7} \ C

         

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Inna Hurry is traveling at 6.8 m/s, when she realizes she is late for an appointment. She accelerates at 4.5 m/s^2 for 3.2 s. Wh
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Answer:

1) v = 21.2 m/s

2) S = 63.33 m

3) s = 61.257 m

4) Deceleration, a = -4.32 m/s²

Explanation:

1) Given,

The initial velocity of Inna, u = 6.8 m/s

The acceleration of Inna, a = 4.5 m/s²

The time of travel, t = 3.2 s

Using the first equation of motion, the final velocity is

                v = u + at

                   = 6.8 + 4.5 x 3.2

                   = 21.2 m/s

The final velocity of Inna is, v = 21.2 m/s

2) Given,

The initial velocity of Lisa, u = 12 m/s

The final velocity of Lisa, v = 26 m/s

The acceleration of Lisa, a = 4.2 m/s²

Using the III equations of motion, the displacement is

                          v² = u² +2aS

                         S = (v² - u²) / 2a

                            = (26² -12²) / 2 x 4.2

                            = 63.33 m

The distance Lisa traveled, S = 63.33 m

3) Given,

The initial velocity of Ed, u = 38.2 m/s

The deceleration of Ed, d = - 8.6 m/s²

The time of travel, t = 2.1 s

Using the II equations of motion, the displacement is

                        s = ut + 1/2 at²

                           =38.2 x 2.1 + 0.5 x(-8.6) x 2.1²

                           = 61.257 m

Therefore, the distance traveled by Ed, s = 61.257 m

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The initial velocity of the car, u = 24.2 m/s

The final velocity of the car, v = 11.9 m/s

The time taken by the car is, t = 2.85 s

Using the first equations of motion,

                         v = u + at

∴                        a = (v - u) / t

                            = (11.9 - 24.2) / 2.85

                            = -4.32 m/s²

Hence, the deceleration of the car, a = = -4.32 m/s²

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