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svetlana [45]
4 years ago
11

A charge of 31.0 μC is to be split into two parts that are then separated by 24.0 mm, what is the maximum possible magnitude of

the electrostatic force between those two parts?
Physics
1 answer:
miskamm [114]4 years ago
8 0

Answer:

1.72 x 10³ N.

Explanation:

When a charge is split equally and placed at a certain distance , maximum electrostatic force is possible.

So the charges will be each equal to

31/2 = 15.5 x 10⁻⁶ C

F = K Q q / r²

= \frac{9\times10^9\times(10.5)^2\times10^{-12}}{(24\times10^{-3})^2}

= 1.72 x 10³ N.

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A 60 kg acrobat is in the middle of a 10 m long tightrope. The center of the rope dropped 30 cm in relation to the ends that are
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Answer:

The tension in each half of the rope, is approximately 4,908.8 N

Explanation:

The mass of the acrobat, m = 60 kg

The length of the rope, l = 10 m

The extent by which the center dropped = 30 cm = 0.3 m

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Weight, W = Mass, m × The acceleration due to gravity, g

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The acceleration due to gravity, g ≈ 9.8 m/s²

∴ The weight of the acrobat, W = 60 kg × 9.8 m/s² ≈ 588 N

The angle the dropped rope makes with the horizontal, θ is given as follows;

θ = arctan((0.3 m)/(5 m)) = arctan(0.06) ≈ 3.434°

At equilibrium, the sum of vertical forces, \Sigma F_y = 0

The vertical component of the tension, T_y, in each half of the rope is given as follows;

T_y = T × sin(θ)

∴ \Sigma F_y = W + T × sin(θ) + T × sin(θ) = W + 2 × T × sin(θ)

Plugging in the values, with θ = arctan(0.06) for accuracy, we get;

588 N + 2 × T × sin(arctan(0.06) = 0

∴ 2 × -T × sin(arctan(0.06) = 588 N

-T= 588 N/(2 × sin(arctan(0.06)) = 4,908.81208 N ≈ 4,908.8 N

The tension in each half of the rope, T ≈ 4,908.8 N.

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Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The minimum velocity of A is  v_A= 4m/s

Explanation:

From the question we are told that

    The length of the string is  L = 1m

     The initial speed of block A is u_A

     The final speed of block A is  v_A = \frac{1}{2}u_A

      The initial speed of block B is u_B = 0

      The mass of block A  is  m_A = 7kg  gh

      The mass of block B is  m_B  = 2 kg

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Since block B at initial is at rest

       m_A u_A  = m_Bv_B + m_A \frac{u_A}{2}

      m_A u_A  - m_A \frac{u_A}{2} = m_Bv_B

          m_A \frac{u_A}{2} = m_Bv_B

  making v_B the subject of the formula

             v_B =m_A \frac{u_A}{2 m_B}

Substituting values

               v_B =\frac{7 u_A}{4}  

This v__B is the velocity at bottom of the vertical circle just at the collision with mass A

Assuming that block B is swing through the vertical circle(shown on the second uploaded image ) with an angular velocity  of v__B' at  the top of the vertical circle  

 The angular centripetal acceleration  would be mathematically represented

                   a= \frac{v^2_{B}'}{L}

Note that  this acceleration would be toward the center of the circle

      Now the forces acting at the top of the circle can be represented mathematically as

         T + mg = m \frac{v^2_{B}'}{L}

    Where T is the tension on the string

  According to the law of energy conservation

The energy at  bottom of the vertical circle   =  The energy at the top of

                                                                                the vertical circle

   This can be mathematically represented as

                 \frac{1}{2} m(v_B)^2 = \frac{1}{2} mv^2_B' + mg 2L

From above  

                (T + mg) L = m v^2_{B}'

Substitute this into above equation

             \frac{1}{2} m(\frac{7 v_A}{4} )^2 = \frac{1}{2} (T + mg) L  + mg 2L  

             \frac{49 mv_A^2}{16}  = \frac{1}{2} (T + mg) L + mg 2L

          \frac{49 mv_A^2}{16}  = T + 5mgL

The  value of velocity of block A needed to cause B be to swing through a complete vertical circle is would be minimum when tension on the string due to the weight of B is  zero

        This is mathematically represented as

                      \frac{49 mv_A^2}{16}  = 5mgL

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            v_A = \sqrt{\frac{80mgL}{49m} }

substituting values

          v_A = \sqrt{\frac{80* 9.8 *1}{49} }

              v_A= 4m/s

     

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3 years ago
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