The propeller of an aircraft is spinning at 10 rev/s when the pilot shuts off the engine. The propeller reduces its angular velocity at a constant 2.0rad/s2 for a time period of 40 s. What is the rotation rate of the propeller in 40 s

One revolution of the propeller corresponds to a rotation of 2π radians, so that the propeller has an initial angular velocity of

10 rev/s = (10 rev/s) • (2π rad/rev) = 20π rad/s

The propeller thus has an angular velocity ω at time t of

ω = 20π rad/s - (2.0 rad/s²) t

so that at t = 40 s, its angular speed is reduced to

20π rad/s - (2.0 rad/s²) (40 s) = (20π - 80) rad/s

Convert this to a rotation rate by dividing this result by 2π :

(20π - 80) rad/s = ((20π - 80) rad/s) • (1/(2π) rev/rad) ≈ -2.73 rev/s

which would suggest that the propeller has started to turn in the opposite direction at a rate of 2.73 rev/s.

i need the answer to this too but idk it ; -;

beta is electron from nucleus

something from nucleus has changed, but charge must be same for cons of charge