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salantis [7]
3 years ago
13

If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of the

curve (a real problem on icy mountain roads). (a) Calculate the ideal speed in (m/s) to take a 100 m radius curve banked at 15°. m/s (b) What is the minimum coefficient of friction needed for a frightened driver to take the same curve at 15.0 km/h?
Physics
1 answer:
PolarNik [594]3 years ago
3 0

Answer:

a. v₁ = 16.2 m/s

b. μ = 0.251

Explanation:

Given:

θ = 15 ° , r = 100 m , v₂ = 15.0 km / h

a.

To determine v₁ to take a 100 m radius curve banked at 15 °

tan θ  = v₁² / r * g

v₁ = √ r * g * tan θ

v₁ = √ 100 m * 9.8 m/s² * tan 15° = 16.2 m/s

b.

To determine μ friction needed for a frightened

v₂ = 15.0 km / h * 1000 m / 1 km * 1h / 60 minute * 1 minute / 60 seg

v₂ = 4.2 m/s

fk = μ * m * g

a₁ = v₁² / r = 16.2 ² / 100 m = 2.63 m/s²

a₂ = v₂² / r = 4.2 ² / 100 m = 0.18 m/s²

F₁ = m * a₁  ,  F₂ = m * a₂

fk = F₁ - F₂   ⇒  μ * m * g = m * ( a₁ - a₂)

μ * g = a₁ - a₂   ⇒  μ = a₁ - a₂ / g

μ = [ 2.63 m/s² - 0.18 m/s² ] / (9.8 m/s²)

μ = 0.251

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A ball is thrown upwards at an unknown speed. in a time of
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Answer:

a)  Initial speed of the ball = 14.45 m/s

b) At height 6 m speed of ball = 9.55 m/s

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Explanation:

a)  We have equation of motion s=ut+\frac{1}{2} at^2, where s is the displacement, u is the initial velocity, t is the time taken and a is the acceleration.

s = 6 m, t = 0.5 seconds, a = acceleration due to gravity value = -9.8m/s^2

 Substituting

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        v^2=14.45^2-2*9.8*6\\ \\ v=9.55m/s

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c) We have equation of motion v^2=u^2+2as, where v is the final velocity

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     0^2=14.45^2-2*9.8*s\\ \\ s=10.65 m

  Maximum height reached = 10.65 m

8 0
3 years ago
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