Q1) Convert 0.00045 kilometers into micrometers, expressing your solutions in scientific notation.
0.00045 km = 0.00045 × (1000 m/1km) x (106 μm/1m) = 450,000 μm = 4.5 × 10⁵ μm ; Note that: 1 km =1000 m; 1m=106 μm
Q2) A car accelerates from 45 km/h to 90 km/h in 12 seconds.
Calculate the acceleration in both m/s² and in g-units (where 1g = 9.8 m/s²).
Initial velocity: vi = 45 km/h = 45 × (1000/3600) m/s = 12.5 m/s ; Final velocity: vf = 90 km/h = 90 × (1000/3600) m/s = 25 m/s ; vf = vi + a t => a= (vf – vi)/t = 1.042 m/s² ; In g-units: a = 1.042 m/s2 × (1g/9.8 m/s2) = 0.106 g
Q3) Two force vectors are applied to the knee joint during walking: Force A of 45 N at 30°north of west, and Force B of 60 N at 45° north of east.
a) Find the x and y components of each force?
b) Find the net force in the x-direction and y-direction.?
c) Calculate the magnitude and direction of the resultant force on the knee joint.
Q4)
F1x=-15 cos(45)o =-10.6 , F1y=15sin45o=10.6; F2x=20, F2y=0; à ΣFx=20-10.6=9.4, ΣFy=10.6 ; a) ΣF = √[(Fx)² + (Fy)²] = √[(9.4)² + (10.6)²] = 14.17 N ; tan α = y-component/x-component = 10.6 /9.4= 1.13; α = 48.3° ;
b) The resultant force is 14.17 N at an angle of 48.3° from the positive x-axis ; ΣFx = ma=> 14.17= 2 × a ; à a = 7.085 m/s²