A hollow, thin-walled cylinder and a solid sphere start from rest and roll without slipping down an inclined plane of length 3 m. The cylinder arrives at the bottom of the plane 2.5 s after the sphere. Determine the angle between the inclined plane and the horizontal.

3.83⁰

Explanation:

First step:

The linear acceleration of a rolling object on an inclined plane is given by the following equation:

where

M = mass of the rolling object

R  = radius of the rolling object

θ = is the angle of the inclination of the plane

I= is the moment if inertia of the object about its centre of mass.

Therefore, let

mass of cylinder = m

mass of the sphere = m

the time taken by the sphere to reach the bottom be t₂

Given that:

length of the inclined plane, S = 3 m

Time taken to reach the bottom t = t₂ + 2.7s

acceleration of a body on the plane:

Value for the hollow cylinder =

Value of I/mR² for solid sphere =

a for cylinder = gsinθ/ (1+1)

                      = gsin θ/2

a for sphere = g sin θ/ (1 + 2/3)

                     5gsin θ/3

from the distance equation s = ut +

3 m =

t =

From equation 1, t₁ = t₂ + 2.5

substituting the values of a for cylinder and a for sphere an solving for θ gives

solving for θ gives θ = 3.83⁰

Therefore, the angle between the inclined plane and horizontal = 3.83⁰

im guessing the west of station b

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .