To solve this exercise it is necessary to apply the concepts related to Work and Kinetic Energy. Work from the rotational movement is described as

In the case of rotational kinetic energy we know that

PART A)
is given in revolutions and needs to be in radians therefore


Replacing in the work equation we have to



PART B) From the torque and moment of inertia it is possible to calculate the angular acceleration and the final speed, with which the kinetic energy can be determined.

Rearrange for the angular acceleration,



From the kinematic equations of angular motion we have,




In this way the rotational kinetic energy would be given by



Answer:
The maximum error is 
Explanation:
From the question we are told that
The length is 
The radius is 
The pressure is 
The rate is 
The viscosity is 
The error in the viscosity is mathematically represented as

Where 
and 
and 
So
![\Delta \eta = \frac{\pi}{8} [ |\frac{r^4}{v} | * \Delta P + | \frac{4 * P * r^3}{v} |* \Delta r + |-\frac{P* r^4}{v^2} |* \Delta v]](https://tex.z-dn.net/?f=%5CDelta%20%20%5Ceta%20%20%3D%20%5Cfrac%7B%5Cpi%7D%7B8%7D%20%5B%20%7C%5Cfrac%7Br%5E4%7D%7Bv%7D%20%20%7C%20%2A%20%20%5CDelta%20%20P%20%20%20%2B%20%20%20%20%7C%20%5Cfrac%7B4%20%2A%20%20P%20%2A%20r%5E3%7D%7Bv%7D%20%20%7C%2A%20%20%5CDelta%20%20r%20%2B%20%20%7C-%5Cfrac%7BP%2A%20r%5E4%7D%7Bv%5E2%7D%20%20%7C%2A%20%20%5CDelta%20%20v%5D)
substituting values
![\Delta \eta = \frac{\pi}{8} [ |\frac{(0.002)^4}{0.5*10^{-9}} | * 1750 + | \frac{4 * 4 *10^{5} * (0.002)^3}{0.5*10^{-9}} |* 0.0002 + |-\frac{ 4*10^{5}* (0.002)^4}{(0.5*10^{-9})^2} |* 0 ]](https://tex.z-dn.net/?f=%5CDelta%20%20%5Ceta%20%20%3D%20%5Cfrac%7B%5Cpi%7D%7B8%7D%20%5B%20%7C%5Cfrac%7B%280.002%29%5E4%7D%7B0.5%2A10%5E%7B-9%7D%7D%20%20%7C%20%2A%20%201750%20%20%20%2B%20%20%20%20%7C%20%5Cfrac%7B4%20%2A%20%204%20%2A10%5E%7B5%7D%20%2A%20%280.002%29%5E3%7D%7B0.5%2A10%5E%7B-9%7D%7D%20%20%7C%2A%20%200.0002%20%2B%20%20%7C-%5Cfrac%7B%204%2A10%5E%7B5%7D%2A%20%280.002%29%5E4%7D%7B%280.5%2A10%5E%7B-9%7D%29%5E2%7D%20%20%7C%2A%20%200%20%5D)
![\Delta \eta = \frac{\pi}{8} [56 + 5120 ]](https://tex.z-dn.net/?f=%5CDelta%20%20%5Ceta%20%20%3D%20%5Cfrac%7B%5Cpi%7D%7B8%7D%20%5B56%20%20%2B%20%205120%20%5D)


Answer:
2,400
Explanation:
Because when we use the formula D=st and we plug in our numbers it should look like D=8x300 . And when we do the math of multiplication it gives us the answer of 2,400 .
Hoped I helped <3
Sleepy~
Answer:
Distance of the point where electric filed is 2.45 N/C is 1.06 m
Explanation:
We have given charge per unit length, that is liner charge density 
Electric field E = 2.45 N/C
We have to find the distance at which electric field is 2.45 N/C
We know that electric field due to linear charge is equal to
, here
is linear charge density and r is distance of the point where we have to find the electric field
So 
r = 1.06 m
So distance of the point where electric filed is 2.45 N/C is 1.06 m