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![a \ student \ is \ able \ to \ lift \ a \ 500(n) \\ weight \ by \ applying \ only \ 100(n) \\ \\ so, \ this \ info \ here, \ we \ simply \ divide \ by \\ how \ much \ this \ kid \ lifted, \ by \ the \ weight \ he/she \\ \ is \ applying. \\ \\ \left[\begin{array}{ccc}\boxed{\boxed{500(n)/100(n)}}\end{array}\right] \\ \\ and \ from \ this,\ your \ answer \ would \ \\ conclude \ to \ be \ (5) \\ \\ \boxed{5} \ would \ be \ your \ answer!](https://tex.z-dn.net/?f=%20a%20%5C%20student%20%20%5C%20is%20%5C%20able%20%5C%20to%20%5C%20lift%20%5C%20a%20%5C%20500%28n%29%20%5C%5C%20weight%20%5C%20by%20%5C%20applying%20%5C%20only%20%20%5C%20100%28n%29%20%5C%5C%20%5C%5C%20so%2C%20%5C%20this%20%5C%20info%20%5C%20here%2C%20%5C%20we%20%5C%20simply%20%5C%20divide%20%5C%20by%20%5C%5C%20how%20%5C%20much%20%5C%20this%20%5C%20kid%20%5C%20lifted%2C%20%5C%20by%20%5C%20the%20%5C%20weight%20%5C%20he%2Fshe%20%5C%5C%20%5C%20is%20%5C%20applying.%20%5C%5C%20%5C%5C%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5Cboxed%7B%5Cboxed%7B500%28n%29%2F100%28n%29%7D%7D%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%20%5C%5C%20and%20%5C%20from%20%5C%20this%2C%5C%20your%20%5C%20answer%20%5C%20would%20%5C%20%5C%5C%20conclude%20%5C%20to%20%5C%20be%20%5C%20%285%29%20%5C%5C%20%5C%5C%20%5Cboxed%7B5%7D%20%5C%20would%20%5C%20be%20%5C%20your%20%5C%20answer%21)
Hope this helps you!
Hope this helps you!
If an 800.0 Kg sports car slows to 8.0 m/s in the forward direction to check out an accident scene and a 1200.0 kg pick-up truck
1 answer:
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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