c^2 = a^2 + b^2 22.1^2= 16.8^2 + b^2 solve for 'b' using pythag. theorem
time to travel horizontal distance (8.7m) = time to fall vertical distance (3.3m) (a) solve for t in the vertical direction (i) initial speed in vertical direction is 0 m/s (ii) d = (initial speed in vertical diection)t + 1/2 g t^2 3.3 m = 0 + 1/2(9.81 m/s) t^2 (iii) solve for...
distance= (5280/1 mi)(2 mi)
force = mass x acceleration mass = 5 kg acceleration = 10 m/s2 plug in the values and you have your answer.
Add all the Ps 4p-13p-p = -10p divide both sides by -10 to isolate P -10P = -150 -10p/-10p = -150/-10 p = 15 check by inserting calculated value into original equation
Ft = m(final velocity) - m(initial velocity) F = ma so equation changes to mat = m(final velocity) - m(initial velocity) plug in values and solve for 'a'
power = force x velociy. Plug the values in and you have your answer.
initial velocity = 78.3 m/s final velocity = 0 m/s distance traveled - 959 m a = ? use (final velocity)^2 = (initial velocity)^2 + 2ad solve equation for 'a'.
use the work-energy theorem determine mass = 16.0N/9.8 m/s^2 initial velocity = 0; so initial KE = 0 final velocity = 15.0 m/s final height = 0; so final PE = 0 initial height = 150 sin 28.0 degrees substitute values in the work-energy theorem expression
initial velocity = 0 m/s final velocity = 4.92 m/s constant acceleration so, (a) average velocity = (initial velocity + final velocity)/2 (b) distance = average velocity x time substitute and calculate