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sattari [20]
3 years ago
12

A uniform ladder of length 10.8 m is leaning against a vertical frictionless wall. The weight of the ladder is 323 N, and it mak

es an angle of 1.16 radians with the floor. A man weighing 734 N climbs slowly up the ladder. When he has climbed to a point that is 7.46 m from the base of the ladder (measured along the ladder), the ladder starts to slip. Find the coefficient of static friction between the floor and the ladder.
Physics
1 answer:
love history [14]3 years ago
8 0

Answer:

0.3625

Explanation:

From the given information:

Consider the equilibrium conditions;

On the ladder, net torque= 0

Thus,

\tau_{net} = 0; and

-fL \s in \theta +m_L g \dfrac{L}{2} cos \theta + mg (7.46\ m) cos \theta = 0

However, by rearrangement;

fL \s in \theta =m_L g \dfrac{L}{2} cos \theta + mg (7.46\ m) cos \theta  \\ \\  \mu(m_L + m) gL \ sin \theta = (323  \ N) ( 10.8 \ meters) \ cos 56^0 + (734 \ N) (7.46 \ m) \ cos  \  66.46^0

\mu= \dfrac{ (323  \ N) ( 10.8 \ m) \ cos 56^0 + (734 \ N) (7.46 \ m) \ cos  \  66.46^0}{\Big [(323 \ N)+(734 \ N) \Big] (10.8 \ m)}

\mathbf{\mu= 0.3625 }

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Two parallel-plate capacitors, 6.0 mF each, are connected in parallel to a 10 V battery. One of the capacitors is then squeezed
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The additional charge transferred to the capacitors by the battery is 60 μC.

The increase in the total charge stored on the capacitors is 60 μC.

<u>Explanation</u>:

                        C = εоA / d

If the separation is halved, then the capacitance will be doubled and according to the equation q = CV, the charge will be doubled too.

  • Initial charge of the capacitor is:

                   q = CV

                      = (6e - 6) \times (10)

                      = 60 μC

      Final charge of the capacitor:

                   q = (2C)V

                      = (2 \times 6e - 6) \times (10)

                      = 120 μC

       additional charge transmitted is:

                               q' = 120 - 60

                                   = 60 μC

  • initial total charge:

                                q_{i} = (C1 + C2) V

                                   = (6 + 6) \times (10)

                                   = 120 μF

        final total charge:

                               q_{f} = (C1 + C2) V

                                   = (2 \times 6 + 6) \times (10)

                                   = 180 μF

        Increase in the charge:

                               q' = 180 - 120

                                   = 60 μC

6 0
3 years ago
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Charge on particle 1, q_1=-7.97\ \mu C=-7.97\times 10^{-6}\ C

Charge on particle 1, q_2=7.75\ \mu C=7.75\times 10^{-6}\ C

The distance between charges, d = 8.09 cm

To find,

The magnitude of the force that one particle exerts on the other.

Solution,

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k is the electrostatic constant

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