Answer:
Explanation:
Given data
Time t=2.5 minutes=150 seconds
Distance A=1600 ft=487.68 m........east
Distance B=2500 ft=762m ........north
To find
Average velocity
Solution
First we need to find the resultant distance magnitude.To find that we apply Pythagorean theorem to find hypotenuse
So
-- As she lands on the air mattress, her momentum is (m v)
Momentum = (60 kg) (5 m/s down) = 300 kg-m/s down
-- As she leaves it after the bounce,
Momentum = (60 kg) (1 m/s up) = 60 kg-m/s up
-- The impulse (change in momentum) is
Change = (60 kg-m/s up) - (300 kg-m/s down)
Magnitude of the change = <em>360 km-m/s </em>
The direction of the change is <em>up /\ </em>.
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 />
