C it reduces the amount of useful work done on objects move it up the ramp
Answer:
41.8m/s^2
Explanation:
Since the dragster starts from rest, initial velocity (u) = 0m/s, final velocity (v) = 25.9m/s, time (t) = 0.62s
From the equations of motion, v = u + at
a = (v - u)/t = (25.9 - 0)/0.62 = 25.9/0.62 = 41.8m/s^2
Answer:
D. 18.60
Explanation:
By the law of conservation, the momentum is neither loss nor gained but instead transfered. When they crash into each other, and stick, they combine to create a total mass of 215 kg. Since the momentum is transfered, the two objects, combined, have a total momentum of 4000 kg-m/s. We know that momentum equals mass times velocity. You then divide 4000 by 215 and get approximately 18.6 m/s
Answer:
largest lead = 3 m
Explanation:
Basically, this problem is about what is the largest possible distance anchorman for team B can have over the anchorman for team A when the final leg started that anchorman for team A won the race. This show that anchorman for team A must have higher velocity than anchorman for team B to won the race as at the starting of final leg team B runner leads the team A runner.
So, first we need to calculate the velocities of both the anchorman
given data:
Distance = d = 100 m
Time arrival for A = 9.8 s
Time arrival for B = 10.1 s
Velocity of anchorman A = D / Time arrival for A
=100/ 9.8 = 10.2 m/s
Velocity of anchorman B = D / Time arrival for B
=100/10.1 = 9.9 m/s
As speed of anchorman A is greater than anchorman B. So, anchorman A complete the race first than anchorman B. So, anchorman B covered lower distance than anchorman A. So to calculate the covered distance during time 9.8 s for B runner, we use
d = vt
= 9.9 x 9.8 = 97 m
So, during the same time interval, anchorman A covered 100 m distance which is greater than anchorman B distance which is 97 m.
largest lead = 100 - 97 = 3 m
So if his lead no more than 3 m anchorman A win the race.
I believe that the answer is C<span />
