Answer:
1. a
2. a [im iffy on this but 95% positive its this]
3. b [walking is a form of aerobics, so i would say b]
Explanation:
Complete Question
The angular speed of an automobile engine is increased at a constant rate from 1120 rev/min to 2560 rev/min in 13.8 s.
(a) What is its angular acceleration in revolutions per minute-squared
(b) How many revolutions does the engine make during this 20 s interval?
rev
Answer:
a

b

Explanation:
From the question we are told that
The initial angular speed is 
The angular speed after
is 
The time for revolution considered is
Generally the angular acceleration is mathematically represented as

=>
=> 
Generally the number of revolution made is
is mathematically represented as

=> 
=> 
Given:
Uniform distributed load with an intensity of W = 50 kN / m on an overhang beam.
We need to determine the maximum shear stress developed in the beam:
τ = F/A
Assuming the area of the beam is 100 m^2 with a length of 10 m.
τ = F/A
τ = W/l
τ = 50kN/m / 10 m
τ = 5kN/m^2
τ = 5000 N/ m^2<span />