On driving your motorcycle in a circle of radius 75 m on wet pavement, the fastest you can go before you lose traction, assuming the coefficient of static friction is 0.20 is 147m/s
Friction helps to maintain the slipping of the vehicle on the road hence lays a very important role.
Maximum velocity of a road with friction is given by the formula,
v = μRg
where, v is the maximum velocity
μ is the coefficient of static friction
R is the radius of the circle road
g is the acceleration due to gravity
Given,
μ = 0.20
R = 75m
g = 9.8m/s²
On substituting the given values in the above formula,
v = 0.20× 75 ×9.8
v = 147m/s
So, the Maximum velocity of the wet road is 147m/s.
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Answer:
Explanation:
The torque of a force is given by:
where
F is the magnitude of the force
d is the distance between the point of application of the force and the centre of rotation of the system
is the angle between the direction of the force and d
In this problem, we have:
, the force
, the distance of application of the force from the centre (0,0)
, the angle between the direction of the force and a
Therefore, the torque is
Answer:
true
Explanation:
free fall is said to be the downward movement of an object under the force of gravity only
Answer:
i) 0.9504
ii) 0.0452
Explanation:
Given data: reliability of hydraulic brakes= 0.96
reliability of mechanical brakes = 0.99
So the probability of stopping the truck = 0.96×0.99= 0.9504
At low speed
case: A works and B does not
= 0.96×(1-0.99) = 0.0096
case2 : B works and A does not
= 0.99×(1-0.96) = 0.0396
Therefore, probality of stopping = 0.0096+0.0396 = 0.0492
