Ahoop with a mass of 2.75 kg is rolling without slipping along a horizontal surface with a speed of 4.5 m/s when it starts down a ramp that makes an angle of 25° below the horizontal. what is the rotational kinetic energy of the hoop after it has rolled 3.0 m down as measured along the surface of the ramp?

Explanation:

given,

mass of hoop = 2.75 Kg

horizontal speed = 4.5 m/s

angle with horizontal = 25°

m g l sin θ = m ( v² - u² )

g l sin θ = ( v² - u² )

9.8 x 3 x sin 25° = ( v² - 4.5² )

12.425 = v² - 20.25

v² = 32.675

v = 5.72 m/s

Rotational

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

according to question, we have said that if the population has twice its initial population, then it becomes

so, time taken to reach this above mentioned population is given by

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.