Answer:
Explanation:
20 km/hr = 5.56 m/s
90 km/hr = 25 m/s
To have just passed C, B must gain first the length of C, then the length of B for a total 400 m
s = s₀ + v₀t + ½at²
if s₀ = 0 at the head of train C when t = 0
for train C, the position in time is
s = 0 + 5.56t + ½(0.2)t²
s = 5.56t + 0.1t²
for train B which must gain 400 m in the same time
s = -400 + 25t + ½(-0.1)t²
s = -400 + 25t - 0.05t²
As both equations equal s, we can set the other sides equal
5.56t + 0.1t² = -400 + 25t - 0.05t²
0.15t² - 19.44t + 400 = 0
quadratic formula positive answer
t = (19.44 + √(19.44² - 4(0.15)(400))) / (2(0.15))
t = 104 s
v = 25 + 104(-0.1) = 14.6 m/s or 52.6 km/hr
Answer:
ANSWERS ARE 100 % correct
Explanation:
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Answer
160,72N
Explanation
W=mg
=(16.4)(9.8)
<span>So we wan't to know what is the velocity after a collision of two railroad cars, one moving to the east and the other moving to the west if m1=2000kg, v1=5m/s and m2=6000kg, v2=3m/s. We can find the solution using the law of conservation of momentum for plastic collisions that states that the momentum must remain constant before (left side of the equation) and after (right side of the equation) the collision: m1*v1+m2*v2=(m1+m2)*v. So now we simply plug in the numbers and get: 2000kg * 5m/s + 6000kg * 3m/s = (2000kg + 6000kg)*v. Now we can write: 10000 kgm/s + 18000 kgm/s = 8000kg * v. To get v, the velocity of both railroad cars after the collision we simply divide both sides of the equation with 8000 kg: so v=3.5m/s to the west. </span>