How Do Forces Affect the Motion of a Doughnut?

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The discussion focuses on calculating the net force and acceleration of a doughnut being pulled by three individuals with different forces and directions. Ted pulls with 5.0N east, Ahmad with 4.0N north, and Alysha with 6.0N northwest, which is equivalent to 45 degrees. Participants clarify that to find the net force, one must resolve the northwest vector into its x and y components, then sum the forces in both directions. The resultant force and its angle relative to the x-axis can be determined using trigonometric functions. The final calculations yield a net force of 8.3N at an angle of 5.2 degrees east of north and an acceleration of 66m/s².
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Question:
Ted pulls a 0.125g doughnut with force 5.0N[E], Ahmad pulls with a force of 4.0N[N], and Alysha pulls with a force of 6.0N[NW].

Find the net force on the doughnut and find the acceleration of the wagon.

Answers: 8.3N[N 5.2* E], 66m/s^2 [N5.2*E]

Ok.. so I am not sure how to get the answers and especially the 5.2 degrees..
I am guessing that the "NW" stated in the question is.. to say 45degrees? Bleh, please help.
 
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NW means 45 degrees.
Now resolve the vector pointing NW along x and y axis.
Then find net x and y components and then the resultant.
The angle between resultant and the x-axis is given by
tanθ = Net Fy/Net Fx
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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