The wheel and axle increases your force. You exert your input force over a long distance and the output force is increased over a shorter distance. (Because the wheel is larger than the axle, the axle rotates and exerts a large output force.) A simple machine with a grooved wheel with a rope or cable wrapped around it.
Answer:
vi = 4.77 ft/s
Explanation:
Given:
- The radius of the surface R = 1.45 ft
- The Angle at which the the sphere leaves
- Initial velocity vi
- Final velocity vf
Find:
Determine the sphere's initial speed.
Solution:
- Newton's second law of motion in centripetal direction is given as:
m*g*cos(θ) - N = m*v^2 / R
Where, m: mass of sphere
g: Gravitational Acceleration
θ: Angle with the vertical
N: Normal contact force.
- The sphere leaves surface at θ = 34°. The Normal contact is N = 0. Then we have:
m*g*cos(θ) - 0 = m*vf^2 / R
g*cos(θ) = vf^2 / R
vf^2 = R*g*cos(θ)
vf^2 = 1.45*32.2*cos(34)
vf^2 = 38.708 ft/s
- Using conservation of energy for initial release point and point where sphere leaves cylinder:
ΔK.E = ΔP.E
0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))
( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))
vi^2 = vf^2 - 2*g*R*( 1 - cos(θ))
vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))
vi^2 = 22.744
vi = 4.77 ft/s
<h3>Given, </h3>
Force,F = 4000 N
Area,a = 50 m²
<h3>We know that, </h3>
Pressure = Force/Area
★ Putting the values in the above formula,we get:
<span>C.
Sample C would be best, because the percentage of the energy
in an
incident wave that remains in a reflected wave from this material
is the
smallest.
The coefficient of absorption is the percentage of incident sound
that's absorbed. So the highest coefficient of absorption results in
the smallest </span><span>percentage of the energy in an
incident wave that remains.
That's what you want. </span>
