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Kazeer [188]
3 years ago
9

The coefficient of static friction for the tires of a race car is 0.950 and the coefficient of kinetic friction is 0.800. The ca

r is on a level circular track of 50.0 m radius on a planet where g = 2.45 m/s² compared to Earth's g=9.8m/s². The maximum safe speed on the track on the planet is ____ times as large as the maximumsafe speed on Earth.a. 0.250b. 0.500c. 1.00d. 2.00e. 4.00
Physics
2 answers:
Arisa [49]3 years ago
4 0

Answer:

speed on planet is 0.5 times of speed on a earth

Explanation:

Given data:

coeeffcient of kinetic friction= \mu+k  =0.8

friction force is given as

f_k = \mu_k Nn -normal force = mg

for max safe speed

fricional force  =  centripetal force

\mu_k N = \frac{mV^2}{R}

\mu_k m g = \frac{mV^2}{R}

v^2 = \mu_k g R

v^2 = 0.8\times 2.45 \times 50v^2 = 98

v = 9.899 m/sec

therefore maximum speed is  [v max} = 9.899 m/sec

\frac{planet\ speed}{ earth\ speed} = \sqrt { \frac{ \mu_k g_p R}{ \mu_k g_e R}}

\frac{V_p}{V_e} = \sqrt{\frac{g_p}{g_e}}

solving for planet speed

v_p = \sqrt{\frac{2.45}{9.8}} v_e

v_p = 0.5 v_e

speed on planet is 0.5 times of speed on a earth

torisob [31]3 years ago
3 0

Answer:

b. 0.500

Explanation:

Given:

  • coefficient of static friction, \mu_s=0.95
  • coefficient of kinetic friction, \mu_k=0.8
  • radius of circular track, r=50\ m
  • gravity on the planet, g=2.45\ m.s^{-2}

<em>For the condition of safe speed the centripetal force must be equal to the static frictional force so that the wheels of the car do not slip in tangential direction or into the curvature of turn.</em>

f=F_c

\mu_s.R=m.\frac{v^2}{r}

\mu_s.m.g=m.\frac{v^2}{r}

\mu_s.g=\frac{v^2}{r} ..........................................(1)

0.95\times 2.45=\frac{v^2}{50}

v=10.7877\ m.s^{-2}

<u>Now putting values for the earth in eq. (1)</u>

0.95\times 9.8=\frac{v_e^2}{50}

v_e=21.5754\ m.s^{-1}

Hence:

\frac{v}{v_e} =\frac{10.7877}{21.5754} =\frac{1}{2}

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