Answer:
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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:
The angle is 
Explanation:
From the question we are told that
The distance of the dartboard from the dart is 
The time taken is 
The horizontal component of the speed of the dart is mathematically represented as
where u is the the velocity at dart is lunched
so
substituting values
=> 
From projectile kinematics the time taken by the dart can be mathematically represented as
=> 
=> 
![\theta = tan^{-1} [0.277]](https://tex.z-dn.net/?f=%5Ctheta%20%20%3D%20%20tan%5E%7B-1%7D%20%5B0.277%5D)
(a) Fx = 1.464 N
(b) Fy = 1.952 N
(c) F(x, y) = 1.464 i + 1.952 j
Given
Mass = 1kg
Acceleration = 2.44 m/s2
Angle with positive X axis = 53°
As we know
F = ma
By substituting value
F= 1×2.44 N
F= 2.44 N
(a) Component of force in X direction
Fx = F Cosθ
Fx = 2.44 Cos(53°)
Fx = 2.44 × 0.60 = 1.464 N
(b) Component of force in Y direction
Fy = F Sinθ
Fy = 2.44 Sin(53°) = 2.44 × 0.80 = 1.952 N
(c) Net force in vector notation
F(x, y) = 1.464 i + 1.952 j
Thus we got net force.
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