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Doss [256]
3 years ago
6

A 8.0 n force acts on a 0.70-kg object for 0.50 seconds. by how much does the object's momentum change (in kg-m/s)? (never inclu

de units in the answer to a numerical question.)
Physics
1 answer:
inysia [295]3 years ago
4 0
For Newton's second law, the force applied to an object is equal to the product between the mass of the object and its acceleration:
F=ma
Rewriting the acceleration as the increment of velocity \Delta v in a time \Delta t: a= \frac{\Delta v}{\Delta t}, F becomes
F=m \frac{\Delta v}{\Delta t}
But given the definition of momentum: p=mv, then m \Delta v represents the momentum change. So we can rewrite F as
F= \frac{\Delta p}{\Delta t}
And re-arranging the formula we can calculate the value of the change in momentum:
\Delta p = F \Delta t=(8.0 N)(0.50 s)=4 kg m/s
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