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nataly862011 [7]
3 years ago
7

A 75.0 kg person is on a Ferris Wheel that has a radius R = 16 m. The tangential velocity of the Ferris Wheel is 8.25 m/s. Calcu

late the period of rotation of the Ferris Wheel. Enter your answer below and include units and round to 3 significant figures. Show your work on paper, scan it, and submit it in the last problem.

Physics
1 answer:
Dovator [93]3 years ago
3 0

Answer:

The period of rotation is

T=8.025s

Explanation:

The person is undergoing simple harmonic motion on the wheel

Given data

mass of the person =75kg

Radius of wheel r=16m

Velocity =8.25m/s

The oscillating period of simple harmonic motion is given as

T=(2*pi)/2=2*pi √r/g

Assuming that g=9.81m/s

Substituting our data into the expression we have

T=2*3.142 √ 16/9.81

T=6.284*1.277

T=8.025s

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What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1
notka56 [123]

Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is  F_{net}=  3.22*10^{-5} \ J

Explanation:

From the question we are told that

    The third charge is  q_3 =  55 nC =  55 *10^{-9} C

    The position of the third charge is  x = -1.220 \ m

     The first charge is q_1 =  -16 nC  =  -16 *10^{-9} \ C

     The position of the first charge is x_1 =  -1.650m

      The second charge is  q_2 =  32 nC  =  32 *10^{-9} C

      The position of the second charge is  x_2 =   0  \ m  

The distance between the first and the third charge is

      d_{1-3} =  -1.650 -(-1.220)

     d_{1-3} = -0.43 \ m

The force exerted on the third charge by the first is  

     F_{1-3} =  \frac{k  q_1 q_3}{d_{1-3}^2}

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

substituting values

      F_{1-3} =  \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}

       F_{1-3} = 4.28 *10^{-5} \ N

 The distance between the second and the third charge is      

  d_{2-3} =  0- (-1.22)

   d_{2-3} =1.220 \ m

The force exerted on the third charge by the first is mathematically evaluated as

       F_{2-3} =  \frac{k  q_2 q_3}{d_{2-3}^2}

substituting values

       F_{2-3} =  \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}

       F_{2-3} =  1.06*10^{-5} N

The net force is

      F_{net} =  F_{1-3} -F_{2-3}

substituting values

    F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}

    F_{net}=  3.22*10^{-5} \ J

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With the switch closed, how does the voltage across the 20-ω resistor compare to the voltage across the 10-ω resistor?
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In a series connection, the voltage is greater across a high-resistance resistor. Therefore, the voltage is much greater for the 20-ohm resistor. However,if it is a parallel circuit, the voltage is just the same for both resistors.
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Two uncharged metal spheres, spaced 25.0 cm apart, have a capacitance of 26.0 pF. How much work would it take to move 12.0 nC of
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Answer:

W=2.76\times 10^{-6}\ J

Explanation:

Given that,

The distance between two spheres, r = 25 cm = 0.25 m

The capacitance, C = 26 pF = 26×10⁻¹² F

Charge, Q = 12 nC = 12 × 10⁻⁹ C

We need to find the work done in moving the charge. We know that, work done is given by :

U=\dfrac{Q^2}{2C}

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