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natali 33 [55]
3 years ago
12

A positive point charge (q = +8.65 x 10-8 C) is surrounded by an equipotential surface A, which has a radius of rA = 2.97 m. A p

ositive test charge (q0 = +4.56 x 10-11 C) moves from surface A to another equipotential surface B, which has a radius rB. The work done by the electric force as the test charge moves from surface A to surface B is WAB = -6.95 x 10-9 J. Find rB.
Physics
1 answer:
ella [17]3 years ago
7 0

Answer:

r_B = 1.88 m

Explanation:

As we know that work done by electric force is given as

W_e = -q\Delta V

so here we know that charge is moving from

r_A = 2.97 m

to another position

so we will have

W_{AB} = \frac{kq_1q_2}{r_A} - \frac{kq_1q_2}{r_B}

-6.95\times 10^{-9} = (9\times 10^9)(8.65\times 10^{-8})(4.56\times 10^{-11})(\frac{1}{2.97} - \frac{1}{r_B})

-0.196 = (\frac{1}{2.97} - \frac{1}{r_B})

r_B = 1.88 m

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Given:

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• riding on a Ferris wheel whose entrance and exit are the same

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Explanation:

Displacement simply means the.change in position of an object. In a situation whereby the initial and final position are thesame, the displacement will be zero.

The statements that describe a situation with a displacement of zero include:

• riding on a Ferris wheel whose entrance and exit are the same

• walking around the block, starting from and ending at the same house

• running exactly one lap around a racetrack

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A curve that has a radius of 90 m is banked at an angle of =10.8∘. If a 1100 kg car navigates the curve at 75 km/h without skidd
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The minimum coefficient of static friction  between the pavement and the tires is 0.69.

The given parameters;

  • <em>radius of the curve, r = 90 m</em>
  • <em>angle of inclination, θ = 10.8⁰</em>
  • <em>speed of the car, v = 75 km/h = 20.83 m/s</em>
  • <em>mass of the car, m = 1100 kg</em>

The normal force on the car is calculated as follows;

F_n = mgcos(\theta)

The frictional force between the car and the road is calculated as;

F_k = \mu_k F_n\\\\F_k = \mu_k mgcos(\theta)

The net force on the car is calculated as follows;

mgsin(\theta) +  \mu_s mgcos(\theta) = \frac{mv^2}{r} \\\\mg(sin\theta \ + \ \mu_s cos\theta)= \frac{mv^2}{r} \\\\g(sin\theta \ + \ \mu_s cos\theta)= \frac{v^2}{r}\\\\sin\theta \ + \ \mu_s cos\theta = \frac{v^2}{rg}\\\\\mu_s cos\theta = sin\theta \  + \ \frac{v^2}{rg}\\\\\mu_s = \frac{sin\theta}{cos \theta} + \frac{v^2}{cos (\theta)rg}\\\\\mu_s = tan(\theta) +   \frac{v^2}{cos (\theta)rg}\\\\\mu_s = tan(10.8) +  \frac{(20.83)^2}{cos(10.8) \times 90 \times 9.8} \\\\\mu_s = 0.19 + 0.5\\\\

\mu_s = 0.69

Thus, the minimum coefficient of static friction  between the pavement and the tires is 0.69.

Learn more here:brainly.com/question/15415163

8 0
3 years ago
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