5. a 500 kg satellite is in a circular orbit at an altitude of 500 km above the earth's surface. because of air friction, the satellite eventually falls to the earth's surface, where it hits the ground with a speed of 2.00 km/s. how much energy was transformed into internal energy by means of air friction?

the energy transformed into internal energy is E fr = 1.57*10¹⁰ J

Explanation:

From energy conservation , the internal energy gained by friction should be equal to the loss of total energy of the satellite

Since the total energy of the satellite can be decomposed into kinetic and potential energy

E fr = ΔE = E₂-E₁

E₁=  K₁+V₁ = 1/2*m*v₁² + m*g*h₁

E₂ = K₂+V₂=  1/2*m*v₂² + m*g*h₂

first, we can choose our reference state so that h₂=0 and h₁=h=500 km

second , we can calculate the approximate the inicial velocity as the velocity required for a stable circular orbit

g = v₁²/(h+R) → v₁² = g*(h+R)

as the velocity diminishes, h diminishes, falling into the earth

assuming the radius of the Earth as R= 6371 km , then

v₁² = g*(h+R) = 9.8 m/s² * (500 km+ 6371 km) *1000 m/km = 6.73 * 10⁷ (m/s)²

replacing values

E₁ = 1/2*m*v₁² + m*g*h₁ = 1/2* 500kg *6.73 * 10⁷ (m/s)² + 500kg* 9.8m/s² * 500 km = 1.67*10¹⁰ J

E₂= 1/2*m*v₂² + m*g*h₂ =  1/2* 500kg *(2000 m/s)² + 0 = 1*10⁹ J

therefore

E fr = ΔE = E₂-E₁ = 1.67*10¹⁰ J - 1*10⁹ J = 1.57*10¹⁰ J  

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.