A motorcyclist is moving in a circular turn of radius 20
m, with uniform speed 36 km/h. The coefficient of
friction between road and tyre is (g = 10 m/s2)
1
0.75
0.5
0.25
Answers
Answered by
0
Answer:
0.5
Explanation:
36km/h=10 m/s
∴v=√μrg
10=√20×10μ
100=200μ
μ=100/200=0.5
Answered by
0
Answer:
ANSWER
Given speed of the vehicle v=36 km/hr=36×
18
5
=10m/s
turning radius r=20m and static friction μ
s
=0.4
The bend angle of the road θ=tan
−1
(
rg
v
2
)=tan
−1
0.5
When the car travels at maximum speed so that it slips upwards and the component μR acts downwards. So,R
1
−mgcosθ−
r
mv
2
sinθ=0-------(A)
and μR
1
+mgsinθ−
r
mv
2
cosθ=0----------(B)
Solving the equation we get,
v
1
=
rg
1+μtanθ
tanθ−μ
=
20×10(
1.2
0.1
)=4.082m/s=14.7 km/h
Hence its possible speed will be between 14.7 and 54 km/h
Explanation:
OH HLO...
BHAI HAI VO MERA...
DUR REH USSE..
SMJHA...
Similar questions