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tigry1 [53]
3 years ago
11

A car enters a horizontal, curved roadbed of radius 50 m. the coefficient of static friction between the tires and the roadbed i

s 0.20. what is the maximum speed with which the car can safely negotiate the unbanked curve?
Physics
2 answers:
liraira [26]3 years ago
8 0

Answer:

Maximum speed, v =  9.89 m/s

Explanation:

It is given that,

Radius of the curve, r = 50 m

The coefficient of static friction between the tires and the roadbed is 0.20, \mu=0.2

To find,

The maximum speed with which the car can safely negotiate the unbanked curve.

Solution,

The net force acting on the car is balanced by its centripetal force such that the maximum speed of the car is given by :

v=\sqrt{\mu rg}

v=\sqrt{0.2\times 50\times 9.8}

v = 9.89 m/s

So, the maximum speed with which the car can safely negotiate the unbanked curve is 9.89 m/s.

UkoKoshka [18]3 years ago
6 0
Previous results tell us the speed (v) is given in terms of the coefficient of friction (k) and the radius of the curve (r) as
  v = √(kgr)
  v = √(0.20·9.8 m/s²·50 m)
  = 7√2 m/s ≈ 9.90 m/s
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Answer:

a) L=0. b) L = 262 k ^   Kg m²/s and c)  L = 1020.7 k^   kg m²/s

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      L = r x p

Bold are vectors; where L is the angular momentum, r the position of the particle and p its linear momentum

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{array}\right] \left[\begin{array}{ccc}i&j&k\\x&y&z\\px&py&pz\end{array}\right]

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a) at launch point r = 0 whereby L = 0

. b) let's find the position for maximum height, we can use kinematics, at this point the vertical speed is zero

   vfy² = voy²- 2 g y

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L = - (m vox) (2.06) k ^

L = - 20 6.36 2.06 k ^

L = 262 k ^   Kg m² / s

The angular momentum is on the z axis

c) At the point of impact, at this point the height is zero and the position on the x-axis is the range

     R = vo² sin 2θ / g

     R = 9² sin (2 45) /9.8

     R = 8.26 m

L = \left[\begin{array}{ccc}i&j&k\\x&0&0\\px&py&0\end{array}\right]

L = - x py k ^

L = - x m voy

L = - 8.26 20 6.36 k ^

L = 1020.7 k^   kg m² /s

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