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Tamiku [17]
3 years ago
14

True or false conceptual physics 2 questions from chapter 25.

Physics
1 answer:
Arlecino [84]3 years ago
3 0

Answer:

Explanation:

1) TRUE; potential difference can be calculated using path integral. Since the electric field is a conservative, the potential difference can be calculated using any path.

2) TRUE; since potential due to a charge is inversely dependent on distance, at infinity the potential will be almost zero.

3) TRUE, W = q.VBA.

4) FALSE; eV is a unit for work (or) energy.

5) TRUE; since the electric force is conservative force. There will be no loss in energy, the decreased potential energy will be coverted to kinetic energy.

6) FALSE; in the direction of electric field the potential decreases.

7) FALSE; equipotential surface is perpendicular to the electric field lines.

8) FALSE; electrostatic potential is scalar quantity. It depends only on the charge and distance from it.

9) FALSE; Inside a conductor the electric field is zero but the electric potential is constant at the value that is at the surface of the conductor.

10) TRUE; as long as the field is being measured outiside the body the bodies act as point charges. So electric fields due to all types of bodies charged identically will be equal.

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A billiard ball of mass m moving with speed v1=1 m/s collides head-on with a second ball of mass 2m which is at rest. What is th
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Answer:

The velocity of mass 2m is  v_B = 0.67 m/s

Explanation:

From the question w are told that

     The mass of the billiard ball A is =m

     The initial speed  of the billiard ball A = v_1 =1 m/s

    The mass of the billiard ball B is = 2 m

    The initial speed  of the billiard ball  B = 0

Let the final speed  of the billiard ball A  = v_A

Let The finial speed  of the billiard ball  B = v_B

      According to the law of conservation of Energy

                 \frac{1}{2} m (v_1)^2 + \frac{1}{2} 2m (0) ^ 2 = \frac{1}{2} m (v_A)^2 + \frac{1}{2} 2m (v_B)^2

              Substituting values  

                \frac{1}{2} m (1)^2  = \frac{1}{2} m (v_A)^2 + \frac{1}{2} 2m (v_B)^2

Multiplying through by \frac{1}{2}m

                1 =v_A^2 + 2 v_B ^2 ---(1)

    According to the law of conservation of Momentum

            mv_1 + 2m(0) = mv_A + 2m v_B

    Substituting values

            m(1)  = mv_A + 2mv_B

Multiplying through by m

           1 = v_A + 2v_B ---(2)

making v_A subject of the equation 2

            v_A = 1 - 2v_B

Substituting this into equation 1

         (1 -2v_B)^2 + 2v_B^2 = 1

         1 - 4v_B + 4v_B^2 + 2v_B^2 =1

          6v_B^2  -4v_B +1 =1

          6v_B^2 -4v_B =0

Multiplying through by \frac{1}{v_B}

          6v_B -4 = 0

            v_B = \frac{4}{6}

            v_B = 0.67 m/s

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