The car will halt at a minimum distance of 98.57 meters on a rainy day having a friction coefficient of 0.109.
Calculation of the minimum distance-
Provided that :
- the speed of the car = 52 km/h
= 52 x 0.278 m/s = 14.45 m/s
- Friction coefficient, μ = 0.109
the regular force exerted on the car,
Along X-direction, the force is
Friction acts in the opposite way along the x-axis
⇒-μ
⇒-μmg = 
⇒
μg
Utilizing the motion equation-
v² = u²+ 2 .a. s
The final speed, v=0 m/s
⇒0² = (14.45)² - 2 μg .s
⇒2 * 0.109 *9.8 *s = (14.45)² = 208.8
⇒s = 208.8 / 2.14 = 97.57 m
It is concluded that the car will halt at a minimum distance of 98.57 meters.
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Answer:
minimum speed v=
Explanation:
Recall the formula for centripetal force;
Centripetal force is the force that is required to keep an object moving in circular part
where;
F=centripetal force
m=mass of object
r=radius of curvature
v= minimum speed
To find minimum speed make v the subject of formula;
v=
Answer:
Weight measurement at poles will be slightly more than at equator.
Explanation:
Firstly earth bulges at equator. This makes the body far from earth center at equator than poles. Secondly, the surface centrifugal forces of earth due to rotation also plays vital role in disturbing the gravity value. The centrifugal forces cancel the effect of gravity more at equator than at poles.These factors combine to create difference of about 0.5% in gravity values at pole and equator.
gravity at pole=9.832 m/s^2
gravity at equator=9.780 m/s^2
corrected question:The heavyweight boxing champion of the world punches a sheet of paper in midair, bringing it from rest up to a speed of 26.5 m/s in 0.044 s . The mass of the paper is 0.003 kg. Part A Find the force of the punch on the paper
Answer:
Force=1.8N
Explanation:
Newtons third law states that in every action there is equal and opposite reaction.
The force of the punch will be the force that moves the paper by a speed of 26.5m/s.
m=0.003kg , v=26.5m/s u=0(the paper is punched from rest) t=0.044s
F=1.8N
