The component of the crate's weight that is parallel to the ramp is the only force that acts in the direction of the crate's displacement. This component has a magnitude of
<em>F</em> = <em>mg</em> sin(20.0°) = (15.0 kg) (9.81 m/s^2) sin(20.0°) ≈ 50.3 N
Then the work done by this force on the crate as it slides down the ramp is
<em>W</em> = <em>F d</em> = (50.3 N) (2.0 m) ≈ 101 J
The work-energy theorem says that the total work done on the crate is equal to the change in its kinetic energy. Since it starts at rest, its initial kinetic energy is 0, so
<em>W</em> = <em>K</em> = 1/2 <em>mv</em> ^2
Solve for <em>v</em> :
<em>v</em> = √(2<em>W</em>/<em>m</em>) = √(2 (101 J) / (2.0 m)) ≈ 10.0 m/s
Friction between two flat surfaces is distinguished as being kinetic friction.
Explanation:
Given that,
The disintegration constant of the nuclide, 
(a) The half life of this nuclide is given by :



(b) The decay equation of any radioactive nuclide is given by :


Number of remaining sample in 4.44 half lives is :


So, 

(c) Number of remaining sample in 14.6 days is :


So, 

Hence, this is the required solution.
Answer:
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