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Masja [62]
3 years ago
12

A 210 Ohm resistor uses 9.28 W of

Physics
1 answer:
stiv31 [10]3 years ago
8 0
  • Resistance=R=210Ohm
  • Power=9.28W=P
  • Current=I

\boxed{\sf P=I^2R}

\\ \sf\longmapsto I^2=\dfrac{P}{R}

\\ \sf\longmapsto I^2=\dfrac{9.28}{210}

\\ \sf\longmapsto I^2\approx0.04

\\ \sf\longmapsto I\approx\sqrt{0.04}

\\ \sf\longmapsto I\approx\sqrt{\dfrac{4}{100}}

\\ \sf\longmapsto I\approx\dfrac{\sqrt{4}}{\sqrt{100}}

\\ \sf\longmapsto I\approx\dfrac{2}{10}

\\ \sf\longmapsto I\approx0.2A

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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

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