A 1200 kg car is travelling qt 20m\s determine the work that must be done on the car by the frictional force of the brakes to st
1 answer:
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Answer:
a)
b) The second runner will win
c) d = 10.54m
Explanation:
For part (a):
For part (b) we will calculate the amount of time that takes both runners to cross the finish line:
Since it takes less time to the second runner to cross the finish line, we can say the she won the race.
For part (c), we know how much time it takes the second runner to win, so we just need the position of the first runner in that moment:
X1 = V1*t2 = 239.46m Since the finish line was 250m away:
d = 250m - 239.46m = 10.54m
Answer:
The final velocity of the second car is 57 m/s south.
Explanation:
This is an elastic collision between two train cars. In this case, the total kinetic energy between the two bodies will remain the same.
The formula to apply is :
where ;
Given in the question that;
Apply the formula as;
{14650*18}+{3825*11} = {14650 *6} + {3825 * v₂f}
263700+42075=87900 + 3825v₂f
305775 =87900 + 3825v₂f
305775-87900 = 3825v₂f
217875=3825v₂f
217875/3825 =v₂f
56.96 = v₂f
<u>57 m/s = v₂f { nearest whole number}</u>