Answer:
The outer diameter of the shaft to 2s.f is 12.65cm
Step-by-step explanation:
We know that the expression for the area of a circular cross section is
A= 
given data
area of cross section= 
bore diameter d= 8cm
from the area of cross section we can solve for the outer diameter D

solving for D we have

Answer: OK, the answer is 28.
Step-by-step explanation: Also there is no remainder for this division problem, you could have the remainder if the numbers are not the same as a multiplication problem, so did you get it?
Given that,
velocity, v = 7.8 ft/s
Radius, r = 1.25 ft
To find,
a. What is the acceleration of the bucket?
b. What is the velocity if the acceleration is 25 ft/sec²?
Solution,
(a) The centripetal acceleration is given by :

Put v = 7.8 ft/s and r = 1.25 ft

(b) Put a = 25 ft/s² to find velocity.

Hence, this is the required solution.
Answer: 7x^2-8x+10
Step-by-step explanation: hope this helps gl :)
thx for the points