An Atwood machine is constructed using two wheels (with the masses concentrated at the rims). The left wheel has a mass of 2.4 kg and radius 24.26 cm. The right wheel has a mass of 2 kg and radius 33.62 cm. The hanging mass on the left is 1.62 kg and on the right 1.09 kg. m3 m1 m4 m2 What is the acceleration of the system

The acceleration of the system is 0.7305 m/s²

Explanation:

Given that:

m₁ = 2.4 kg , m₂ = 2 kg, m₃ = 1.62 kg, m₄ = 1.09 kg

Torque (τ) = mra

For left pulley:

T₁r₁ - T₂r₁ = m₁r₁a

Dividing through by r₁

T₁ - T₂ = m₁a  1)

For right pulley:

T₂r₂ - T₃r₂ = m₂r₂a

Dividing through by r₂

T₂ - T₃ = m₂a  2)

Also

m₃g - T₁ = m₃a ... 3)

T₃ - m₄g = m₄a  ... 4)

adding equations 1, 2, 3 and 4 to get:

(m₁ + m₂ + m₃ + m₄)a = (m₃ - m₄)g

(2.4 + 2 + 1.62 + 1.09)a = (1.62 - 1.09)9.8

7.11a = 5.194

a = 5.194 / 7.11 = 0.7305

a = 0.7305 m/s²

The acceleration of the system is 0.7305 m/s²

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.

neither accurate nor precise.