# Ings19

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• Birthday 12/04/1985

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1. ## Acceleration + distance

What do you guys think?
2. ## Acceleration + distance

Thanks, for part A I used; final velocity= initial velocity+ rate if acceleration x time = 14m/s Distance during acceleration and deceleration = (14/2) x 20s = 140m Distance= uniform velocity x time taken = 14x300s= 4200m so answer to part a come to 140+4200+140=4480m then part b 40tonnes =40,000kg F=mxa = 40,0000x0.7 =28,000n =28Kn
3. ## Acceleration + distance

A mass of 40tonnes is accelerated from rest for 20s then runs uniformly for 5 mins and is finally brought to rest. Assuming acceleration and deceleration are 0.7m/s2 calculate; a) total distance travelled) b)total force required during this period (ignoring friction) for a I have calculated 4480m for b I have calculated 28kn. (F=m x a = 40000 x0.7)
4. ## Potential energy and kinetic energy

Hi, the question is...a body of mass 200kg is raised to a height of 5m before being allowed to fall freely. Calculate the potential energy and kinetic energy; a) before it starts to fall when it reaches the ground. i have calculated for P.E@5m = PE=m x g x h = 200x10x5 = 10kJ P.E@0m= m x g x h = 200x10x0= 0J k.e@5m= ke= (m x v2)/2 = (200x0)/2 = 0J k.e @0m= (m x v2)/2 = (200x10sq)/2 = 10KJ
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