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Kitty [74]
3 years ago
5

On a different day the same car enters a 420-m radius horizontal curve on a rainy day when the coefficient of static friction be

tween its tires and the road is 0.250. What is the maximum speed at which the car can travel around this curve without sliding?
Physics
1 answer:
Inessa05 [86]3 years ago
7 0

Answer:

centripetal force points towards the center of the circle, and force friction points the opposite way. Set the two equal to each other

Force Normal * coefficient of static friction = mass*velocity^2/radius

(mg)Us = mv^2/r

mass isn't given in the problem because it cancels on both sides.

9.8Us = V^2/r

9.8(.600) = v^2/300

v^2 = 1764

v = 42 m/s

the man

Oct 30, 2015

A car enters a 300-m radius flat curve on a rainy day when the coefficient of static friction between its tires and the road is 0.600. What is the maximum speed which the car can travel around the curve without sliding?

m/s 33.1 0

m/s 29.6 0

m/s 42

m/s 24.8

Explanation:

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Answer:

a)  v = √(v₀² + 2g h),    b)      Δt = 2 v₀ / g

Explanation:

For this exercise we will use the mathematical expressions, where the directional towards at is considered positive.

The velocity of each ball is

ball 1. thrown upwards vo is positive

        v² = v₀² - 2 g (y-y₀)

in this case the height y is zero and the height i = h

        v = √(v₀² + 2g h)

ball 2 thrown down, in this case vo is negative

         v = √(v₀² + 2g h)

The times to get to the ground

ball 1

         v = v₀ - g t₁

         t₁ = \frac{v_{o}  - v }{ g}

ball 2

         v =  -v₀ - g t₂

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From the previous part, we saw that the speeds of the two balls are the same when reaching the ground, so the time difference is

       Δt = t₂ -t₁

       Δt = \frac{1}{g} \ [(v_{o} - v)  - ( - v_{o}  - v) ]

       Δt = 2 v₀ / g

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