Net Force on a 850 kg Car Moving Right at 1.44 m/s

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For an 850 kg car moving at a constant velocity of 1.44 m/s to the right, the net force acting on it is zero. This is because, according to Newton's first law, an object in motion at a constant velocity has no acceleration. Therefore, regardless of the direction of movement, if the velocity remains constant, the net force remains unchanged. Understanding Newton's laws clarifies that constant velocity implies no net force is required to maintain that motion. The discussion emphasizes the importance of these fundamental principles in solving physics problems.
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An 850 kg car is moving to the right at 1.44 m/s. What is the net force on the car ?

I am totally off on this problem ! Can someone help ?
And in the problem above, it is different when the car moving left ?

Thanks
 
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The problem appears incomplete. I am assuming that it means a "constant velocity" to the right at 1.44 m/s.

No matter what it is, or which way it's going, or how fast it's going...

if an object is going at a constant velocity, what is its acceleration, and what's the net force on it?
 
Is that the acceleration is 0, and i don't really know what is net force is ? That why i asked
 
Go read Newton's first and second law.
 
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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