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Montano1993 [528]
3 years ago
13

A bicyclist of mass 60kg supplies 340W of power while riding into a 5 m/s head wind. The frontal area of the cyclist and bicycle

together is 3.7ft2(0.344 m2), the drag coefficient is 0.88, and the coefficient of rolling resistance is 0.007. Determine the speed of the cyclist.Express your answer in mph. (Hint: Lecture Not
Physics
1 answer:
NikAS [45]3 years ago
7 0

Answer:

87.1 mph

Explanation:

We are given that

Mass,m=60 kg

Power,P=340 W

Speed,v=5 m/s

Area,A=0.344 m^2

Drag coefficient,C_d=0.88

Coefficient of rolling resistance,\mu_r=0.007

Friction force,f=\mu_rmg=0.007\times 60\times 9.8=4.1 N

Where g=9.8 m/s^2

Let speed of cyclist=v'

Drag force,F_d=\frac{1}{2}\rho_{air}AC_dv^2

Density of air,\rho_{air}=1.225 kg/m^3

F_d=\frac{1}{2}\times 1.225\times 0.344\times 0.88(5)^2=4.635N

Power,P=(F_d+f)\times v'

340=(4.1+4.635) v'=8.735v'

v'=\frac{340}{8.735}=38.9m/s

v'=87.1 mph

1 m=0.00062137 miles

1 hour=3600 s

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solution:

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A disk of a radius 50 cm rotates at a constant rate of 100 rpm. What distance in meters will a point on the outside rim travel d
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Answer:

A point on the outside rim will travel 157.2 meters during 30 seconds of rotation.

                         

Explanation:

We can find the distance with the following equation since the acceleration is cero (the disk rotates at a constant rate):

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Where:

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t: is the time = 30 s  

The tangential speed can be found as follows:

v = \omega*r

Where:

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Now, the distance traveled by the disk is:

d = v*t = 5.24 m/s*30 s = 157.2 m

Therefore, a point on the outside rim will travel 157.2 meters during 30 seconds of rotation.

I hope it helps you!

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And you can get it using the following equation:

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Where :

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