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Rufina [12.5K]
3 years ago
11

A Honda Hawk motorcycle and its rider with a combined mass of 450 kg travels around a curve of radius 106 m with a speed of 18 m

/s. Calculate the coefficient of static friction of the pavement if they negotiate the curve without skidding. the normal force is 4410.
Physics
1 answer:
umka21 [38]3 years ago
7 0

Answer: coefficient of static friction

= 0.31

Explanation: Since they negotiate the curve without skidding, the frictional force (F1) equals the centripetal force (F2).

F1= uN

F2 = M*(v²/r)

M is the combined mass 450kg

V is the velocity 18m/s

r is the radius 106m

N is the normal reaction 4410N

u is the coefficient of static friction

Making u subject of the formula we have that,

u = {450*(18²/106)} /4410

=1375.47/4410

=0.31

NOTE: coefficient of friction is dimensionless. It as no Unit.

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Answer:

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b) The exit area of the nozzle is 23.6 cm².

Explanation:

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Where:

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\rho_{i}: is the density of entrance = 2.21 kg/m³

\dot{m} = \rho_{i} v_{i}A_{i} = 2.21 \frac{kg}{m^{3}}*20 \frac{m}{s}*60 cm^{2}*\frac{1 m^{2}}{(100 cm)^{2}} = 0.27 kg/s  

Hence, the mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle can be found with the Continuity equation:

\rho_{i} v_{i}A_{i} = \rho_{f} v_{f}A_{f}

0.27 kg/s = 0.762 kg/m^{3}*150 m/s*A_{f}

A_{f} = \frac{0.27 kg/s}{0.762 kg/m^{3}*150 m/s} = 0.00236 m^{2}*\frac{(100 cm)^{2}}{1 m^{2}} = 23.6 cm^{2}

Therefore, the exit area of the nozzle is 23.6 cm².

I hope it helps you!                                                                  

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<h3>♫ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ♫</h3>

➷ A normal atom has the same amount of electrons and protons, making it neutral. An atom develops a positive charge when it loses an electron(s). Once it loses an electron(s), there would now be more protons that electrons.

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➶ Hope This Helps You!

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