Explain Force of Thrust Equation: 6N/3.48s

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The discussion centers on the Force of Thrust equation, specifically how to interpret the calculation of 6N divided by 3.48s resulting in 1.7 N. Participants clarify that the equation represents the change in momentum over time, where dividing Newtons by seconds cancels out the time unit, leaving Newtons as the final unit. The correct formula for Force of Thrust is presented as the change in momentum divided by the change in time. Some users express frustration over the initial question being redundant, suggesting that reviewing previous textbook pages could provide clarity. Overall, the conversation emphasizes understanding the relationship between force, momentum, and time in the context of thrust.
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Please Explain this?!

hey, ok so i was going tthrough my textbook and following some examples hwne i cam up to this:

Force of Thrust = Ft/ delta T

= 6N / 3.48s

=1.7 N

but it doesn't say what the equation means. can someone please help me and explain what is happeing.
 
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A little more information regarding the problem would be helpful. However, how can one divide Newtons by time and wind up with Newtons?
 


Have you covered momentum yet? I don't remember how to calculate force of thrust without using momentum, so I hope this is helpful to you.

The equation for Force of Thrust I know, is
F(thrust) = ((m * V)2 - (m * V)1) / (t2 - t1)

This can also be written as
F(thrust) = Δp / Δt
Force of thrust = change in momentum / change in time

Which would give you
N = N*s/s
meaning your seconds would cancel to give the correct units.
 


the doth prevous questins are pretty good. But i don't understand why did u posted a quastion that is already answered.. If you don't understand it just read the 3-4 prevous pages of your book! That is not so difficult (sorry about my english and wrong words sometimes,,, ist because I am writing from my mobile)
 
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