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expeople1 [14]
3 years ago
8

How are distance and time related when describing motion?

Physics
1 answer:
Bingel [31]3 years ago
3 0
an object moves along a straight line, the distance travelled can be represented by a distance-time graph. In a distance-time graph, the gradient of the line is equal to the speed of the object. The greater the gradient (and the steeper the line) the faster the object is moving
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With more particles there will be more collisions and so a greater pressure. The number of particles is proportional to pressure, if the volume of the container and the temperature remain constant. ... This happens when the temperature is increased.

Explanation:

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In which situation is the acceleration of the car negative?
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A 3-column table with 4 rows. The first column titled substances has entries aluminum, zinc, chromium, nickel. The second column
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wo parallel-plate capacitors C1 and C2 are connected in parallel to a 12.0 V battery. Both capacitors have the same plate area o
tresset_1 [31]

Complete Question

Two parallel-plate capacitors C1 and C2 are connected in parallel to a 12.0 V battery. Both capacitors have the same plate area of 5.30 cm2 and plate separation of 2.65 mm. However, the first capacitor C1 is filled with air, while the second capacitor C2 is filled with a dielectric that has a dielectric constant of 2.10.

(a) What is the charge stored on each capacitor

 (b)  What is the total charge stored in the parallel combination?

Answer:

a

   i    Q_1 =  2.124 *10^{-11} \  C

   ii    Q_2 =  4.4604 *10^{-11} \ C

b

  Q_{eq} = 6.5844 *10^{-11} \ C

Explanation:

From the question we are told that

   The  voltage of the battery is  V  = 12.0  \ V

    The  plate area of each capacitor is  A  =  5.30 \ cm^2  =  5.30 *10^{-4} \ m^2

    The  separation between the plates is  d =  2.65 \ mm =  2.65 *10^{-3} \ m

     The permittivity of free space  has a value  \epsilon_o  =  8.85 *10^{-12} \  F/m

     The  dielectric constant of the other material is  z =  2.10

The  capacitance of the  first capacitor is mathematically represented as

       C_1  =  \frac{\epsilon  *  A }{d }

substituting values

        C_1  =  \frac{8.85 *10^{-12 } *   5.30 *10^{-4} }{2.65 *10^{-3} }

       C_1  =  1.77 *10^{-12} \  F

The  charge stored in the first capacitor is  

       Q_1 =  C_1 *  V

substituting values

        Q_1 =  1.77 *10^{-12} * 12

       Q_1 =  2.124 *10^{-11} \  C

The capacitance of the second  capacitor is mathematically represented as

       C_2  =  \frac{ z * \epsilon  *  A }{d }

substituting values

       C_1  =  \frac{  2.10 *8.85 *10^{-12 } *   5.30 *10^{-4} }{2.65 *10^{-3} }

       C_1  =  3.717 *10^{-12}  \ F

The  charge stored in the second capacitor is  

      Q_2 =  C_2 *  V

substituting values

     Q_2 = 3.717*10^{-12} *  12

     Q_2 =  4.4604 *10^{-11} \ C

Now  the total charge stored in the parallel combination is mathematically represented as

     Q_{eq} =  Q_1 + Q_2

substituting values

    Q_{eq} =  4.4604 *10^{-11} + 2.124*10^{-11}

     Q_{eq} = 6.5844 *10^{-11} \ C

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