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Alenkasestr [34]
3 years ago
11

What is the smallest radius of an unbanked (flat) track around which a bicyclist can travel if her speed is 29.0 km/h and the co

efficient of static friction between tires and track is 0.320? The bicyclist, bicycle and tires are all to the right of the center of the track. Treat the bicyclist, bicycle and tires as one object.
Physics
1 answer:
bazaltina [42]3 years ago
5 0

Answer:

Explanation:

Let the radius of track required be r.

Centripetal force will be provided by frictional force which will be equal to

m v²/ r

Frictional force = mg x μ

So

m v² /r = mg μ

r = v² / μ g =

  v = 29 km /h = 8.05 m /s

r =( 8.05 x 8.05 ) /( .32 x 9.8 ) = 20.66 m

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Gnesinka [82]

Answer: c

Explanation:

The way to check which one is the correct one is to simply multiply and see if there are the same number of atoms in both sides for each element.

a. 2×2 atoms of Al ≠ 3×1 atoms of Al

2×3 atoms of O = 3×2 atoms of O

BOTH MUST BE EQUAL FOR IT TO BE ADJUSTED!!!!!

b. 3×2 atoms of Al ≠ 3×1 atoms of Al

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c. 2×2 atoms of Al = 4×1 atoms of Al

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d. 2×2 atoms of Al ≠ 1×1 atoms of Al

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Answer:

x=22.65m

Explanation:

We have an uniformly accelerated motion, with a negative acceleration. Thus, we use the kinematic equations to calculate the distance will it take to bring the car to a stop:

v_f^2=v_0^2-2ax\\\frac{0^2-v_0^2}{-2a}=x\\x=\frac{v_0^2}{2a}

The acceleration can be calculated using Newton's second law:

\sum F_x:F_f=ma\\\sum F_y:N=mg

Recall that the maximum force of friction is defined as F_f=\mu N. So, replacing this:

\mu N=ma\\\mu mg=ma\\a=\mu g\\a=0.901(9.8\frac{m}{s^2})\\a=8.83\frac{m}{s^2}

Now, we calculate the distance:

x=\frac{v_0^2}{2a}\\x=\frac{(20\frac{m}{s})^2}{2(8.83\frac{m}{s^2})}\\x=22.65m

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