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8090 [49]
3 years ago
9

You're driving a vehicle of mass 1350 kg and you need to make a turn on a flat road. The radius of curvature of the turn is 71 m

. The maximum horizontal component of the force that the road can exert on the tires is only 0.23 times the vertical component of the force of the road on the tires (in this case the vertical component of the force of the road on the tires is mg, the weight of the car, where as usual g = +9.8 N/kg, the magnitude of the gravitational field near the surface of the Earth). The factor 0.23 is called the "coefficient of friction" (usually written "", Greek "mu") and is large for surfaces with high friction, small for surfaces with low friction.
(a) What is the fastest speed you can drive and still make it around the turn? Invent symbols for the various quantities and solve algebraically before plugging in numbers.
maximum speed =_______________ m/s
Physics
1 answer:
sergey [27]3 years ago
5 0

Answer:

v=12.65\ m.s^{-1}

Explanation:

Given:

  • mass of vehicle, m=1350\ kg
  • radius of curvature, r=71\ m
  • coefficient of friction, \mu=0.23

<u>During the turn to prevent the skidding of the vehicle its centripetal force must be equal to the opposite balancing frictional force:</u>

m.\frac{v^2}{r} =\mu.N

where:

\mu= coefficient of friction

N= normal reaction force due to weight of the car

v= velocity of the car

1350\times \frac{v^2}{71} =0.23\times (1350\times 9.8)

v=12.65\ m.s^{-1} is the maximum velocity at which the vehicle can turn without skidding.

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1.a bag is dropped from a hovering helicopter. the bag has fallen for 2 s. what is the ball's velocity at the instant its hittin
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Recall this kinematics equation:

Vf = Vi + aΔt

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Given values:

Vi = 0m/s (you assume this because the bag is dropped, so it falls starting from rest)

a is 9.81m/s² (this is the near-constant acceleration of objects near the surface of the earth)

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Plug in the values and solve for Vf:

Vf = 0 + 9.81×2

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2. The height of the helicopter.

Recall this other kinematics equation:

d = ViΔt + 0.5aΔt²

d is the distance traveled by the object, Vi is the initial velocity, a is the acceleration, and Δt is the time elapsed.

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Plug in the values and solve for d:

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3. Time of the bag's fall and its velocity immediately before hitting the ground... if it started falling at 2m/s

Reuse the equation from question 2:

d = ViΔt + 0.5aΔt²

Given values:

d = 19.6m (height of the helicopter obtained from question 2)

Vi = 2m/s

a = 9.81m/s² (acceleration due to earth's gravity)

Plug in the values and solve for Δt:

19.6 = 2Δt + 0.5×9.81Δt²

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Δt = 1.8s, Δt = −2.2s

The formula gives us 2 possible answers for Δt but within the situation of our problem, only the positive value makes sense. Reject the negative value.

Δt = 1.8s

Now we can use this new value of Δt to get the velocity before hitting the ground:

Vf = Vi + aΔt

Given values:

Vi = 2m/s

a = 9.81m/s²

Δt = 1.8s (result from previous question)

Plug in the values and solve for Vf:

Vf = 2 + 9.81×1.8

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