Answer:
65.625 m/s
Explanation:
c = Speed of light = 3×10⁸ m/s
Relative speed
Relative speed = 35.625 m/s
Velocity of speeder = Velocity of police car + Relative speed
⇒Velocity of speeder = 30 + 35.625 = 65.625 m/s
∴ Velocity of speeder is 65.625 m/s
Answer:
Average velocity is 3.93 cm/s
Root mean square velocity is 4.79 cm/s
Velocity peak to peak is 6.28 cm/s
Explanation:
Speed of 2 particles = 5.3 cm/s
Speed of 4 particles = 1.4 cm/s
Speed of 6 particles = 7.14 cm/s
Speed of 8 particles = 1.52 cm/s
Speed of 2 particles = 7.68 cm/s
Average velocity is 3.93 cm/s
Root mean square velocity is 4.79 cm/s
Velocity peak to peak is 6.28 cm/s
The magnitude of the total friction force exerted by the road on the tires at B will be 919.46 lb.
<h3>What is the friction force?</h3>It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. Its unit is Newton (N).
Mathematically, it is defined as the product of the coefficient of friction and normal reaction.
The given data in the problem is;
The weight is,W= 3,220 lb
The speed is,u= 60 mi/hr
The reducing speed is,v= 45 mi/hr
The distance traveled is,d= 300 ft
The radius of curvature of the path of the car at B is,R= 600 ft.
1 mile = 5280 ft
From the Newtons' equation of motion;
The tangential accelerations are;
The force is found as;
The normal force is;
The net or the total friction force exerted by the road on the tires at B. is found as;
Hence, the magnitude of the total friction force exerted by the road on the tires at B will be 919.46 lb.
To learn more about the friction force, refer to the link;
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