How many revolutions does the object make during the first 3.9 s?

The area which is located under the curve gives us the angle in radians; we divide this by 2π to get units in revs (revolutions).

So for 0 ≤ t ≤ 2,

number of revs = 20rad/s * 2s * 1rev/2π rads = 6.37 revs 

So for 2 ≤ t ≤ 3.9,

number of revs = 10rad/s * 1.9s * 1rev/2πrads = 3.02 revs 

Therefore the total is about:

6.37 revs + 3.02 revs = 9.39 revs ≈ 9.4 revs

r=rho l/a where a is the area, l is the length and rho is resistivity so the greater the area the lesser the resistance thats why the earth pin is thicker bcos current follow the shortest path, so less resistive .

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer