The IMA of the pulley shown is 2.
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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<h2>
Answer:53.63
</h2>
Explanation:
The equations of motion used in this question is 
When a object is projected horizontally from a sufficiently height,the x-component of acceleration remains zero because there is no force that drags the object in x direction.
But,due to gravity,the object accelerates downward at a rate of
.
In X-Direction,
Given that initial velocity=
=
Using
,

In Y-Direction,
Given that initial velocity=
=
Using
,



Answer:
500 m
Explanation:
t = Time taken
u = Initial velocity = 50 m/s
v = Final velocity = 0
s = Displacement
a = Acceleration = -2.5 m/s²
Equation of motion

Time taken by the train to stop is 20 seconds

∴ The engineer applied the brakes 500 m from the station
Answer:
- 1.07 ft
Explanation:
V1 = (-5, 7, 2)
V2 = (3, 1, 2)
Projection of v1 along v2, we use the following formula
=\frac{\overrightarrow{V1}.\overrightarrow{V2}}{V2}
So, the dot product of V1 and V2 is = - 5 (3) + 7 (1) + 2 (2) = -15 + 7 + 4 = -4
The magnitude of vector V2 is given by
= 
So, the projection of V1 along V2 = - 4 / 3.74 = - 1.07 ft
Thus, the projection of V1 along V2 is - 1.07 ft.
so we need to find the direction of v2