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A fancart of mass 0.8 kg initially has a velocity of < 0.9, 0, 0 > m/s. Then the fan is turned on, and the air exerts a constant force of < -0.3, 0, 0 > N on the cart for 1.5 seconds. What is the change in momentum of the fancart over this 1.5 second interval? = < , 0, 0 > kg

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MechEngineer

Answer:

<-0.45, 0, 0> kgm/s

Explanation:

The change in momentum would equal to the impulse generated by force F = <-0.3, 0, 0> over time duration of Δt = 1.5 second. The impulse is defined as the product of force F and the duration Δt

[tex]\Delta p = F \Delta t = (-0.3) * 1.5 = -0.45 kgm/s[/tex]

So the change in momentum is <-0.45, 0, 0> kgm/s

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