Answer:
<em>The velocity after the collision is 2.82 m/s</em>
Explanation:
<u>Law Of Conservation Of Linear Momentum
</u>
It states the total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is
P=mv.
If we have a system of two bodies, then the total momentum is the sum of the individual momentums:
If a collision occurs and the velocities change to v', the final momentum is:
Since the total momentum is conserved, then:
P = P'
Or, equivalently:
If both masses stick together after the collision at a common speed v', then:
The common velocity after this situation is:
There is an m1=3.91 kg car moving at v1=5.7 m/s that collides with an m2=4 kg cart that was at rest v2=0.
After the collision, both cars stick together. Let's compute the common speed after that:
The velocity after the collision is 2.82 m/s
Answer:
Hello there, please see explanations for step by step answer.
Explanation:
Radius 6 ft and
Height 18 ft is filled to a height of 11 ft of a liquid weighing 64.4 lb divided by ft cubedlb/ft3.
How much work will it take to pump the contents to the rim.
See attached documents for clear solvings and further step by step explanations
Weight = (mass) x (gravity)
= 200 Newtons.
(About 44 pounds. You're very skinny.)
Answer:
s = 3 m
Explanation:
Let t be the time the accelerating car starts.
Let's assume the vehicles are point masses so that "passing" takes no time.
the position of the constant velocity and accelerating vehicles are
s = vt = 40(t + 2) cm
s = ½at² = ½(20)(t)² cm
they pass when their distance is the same
½(20)(t)² = 40(t + 2)
10t² = 40t + 80
0 = 10t² - 40t - 80
0 = t² - 4t - 8
t = (4±√(4² - 4(1)(-8))) / 2(1)
t = (4± 6.928) / 2 ignore the negative time as it has not occurred yet.
t = 5.464 s
s = 40(5.464 + 2) = 298.564 cm
300 cm when rounded to the single significant digit of the question numerals.
