Ahigh-speed drill reaches 2300 rpm in 0.54 s. a) what is the drill's angular acceleration? b. through how many revolutions does it turn during this first 0.50 s?

The angular acceleration of the drill is 446.1 rad/s².

The number of revolutions of the drill in 0.5 second is 19.2 revolutions.

The given parameters;

The angular speed in radian per second is calculated as follows;

The angular acceleration of the drill is calculated as follows;

The number of revolutions of the drill in 0.5 second is calculated as follows;

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(A)Angular acceleration,

(B) Number of revolution is 8.87.

Explanation:

Given ,time taken to reach 2300 rpm speed from 0 is ,t = 0.54 s.

2300 rpm = 240.85 rad/s

(A) Applying equation of motion to find the drill's angular acceleration

(B) Let there are number of revolutions it turn during this first 0.50 s. It can be calculated using second equation of motion as :

Radian can be converted to revolution as :

Therefore, this is the required solution.

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14

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