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lys-0071 [83]
3 years ago
10

a motorcycle traveling on the highway at a velocity of 120 kilometre-per-hour passes a car traveling at a velocity of 90 kilomet

re-per-hour from the point of view of a passenger on the car what is the velocity of a motorcycle​
Physics
1 answer:
musickatia [10]3 years ago
5 0

Answer:

The relative velocity of the motorcycle to a passenger in the car is 30 km/h

Explanation:

The question relates to the principle of relative velocity and reference frames

The given parameters are;

The velocity of the motorcycle, v₁ = 120 km/h

The velocity of the car, v₂ = 90 km/h

The relative velocity of an object X with regards to another object Y is the velocity the object X will seem to be moving with to an observer in the rest frame of object Y written as \underset{v}{\rightarrow}_{X|Y} = \underset{v}{\rightarrow}_{X} - \underset{v}{\rightarrow}_{Y}

Therefore, the relative velocity of the motorcycle to the car is v_{1|2} = v₁ -  v₂, which give;

v_{1|2}  = 120 km/h - 90 km/h = 30 km/h

The relative velocity of the motorcycle to a passenger in the car = 30 km/h.

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4 years ago
Part 1 :
dybincka [34]

Answer:

1. 218.55 N

2. 30.96^{o}

3. 2.1 m/s^{2}

Explanation:

Part 1;

Net force F=mg sin \theta where m is mass, g is gravitational force and \theta is the angle of inclination

F= 46*9.8*sin 29^{o}= 218.55N

Frictional force, F_{r} is given by

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Part 2.

Using the relationship that

Frictional force F_{s} = \mu_{s} mg cos \theta

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\mu_{s}= \frac {sin \theta}{cos \theta}

\mu_{s}= tan \theta

The maximum angle of inclination \theta = tan^{-1} \mu_{s}

\theta = tan^{-1} (0.6)

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Part 3:

Net force on the object is given by

ma = mg sin 38 - \mu_{k} mg cos 38 where \mu_{k} is the coefficient of kinetic friction

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                = 2.1m/s^{2}

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