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:
0.903 seconds
Explanation:
To find how many seconds the acorn fall, we can use the formula for distance travelled with constant acceleration:
D = Vo*t + a*t^2/2,
where D is the distance travelled, Vo is the inicial speed, t is the time and a is the acceleration.
In our problem:
Vo = 0,
a = g = 9.81 m/s2,
D = 4 meters.
So, we can solve the equation to find the time:
4 = 0*t +9.81*t^2/2
4.905*t^2 = 4
t^2 = 4/4.905 = 0.8155
t = 0.903 seconds
Weight = m times g = 5.23 times 8.83 = 46.18 N
Answer:8.968 N-m
Explanation:
Given
Length of arm=0.152 m
Downward force=118 N
angle made by arm with vertical
Force can be divided into two components
It's sin component will contribute towards torque while cos component will not contibute


T=8.968 N-m