Solving a Vector Problem in the xy Plane

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To solve the vector problem, the x component can be determined using the Pythagorean theorem, yielding two possibilities: x = 40.0 or x = -40.0. Given that the x component is positive, the vector is (40.0, -60.0). To achieve a resultant vector of 70.0 units pointing in the -x direction, a vector with components (-70.0, 0) must be added to the original vector. The equation linking the components x, y, and magnitude r is r = √(x² + y²). This discussion emphasizes the application of vector addition and the relationship between components and magnitude in the xy plane.
jpodo
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Here is the problem I can't solve:

You are given a vector in the xy plane that has a magnitude of 80.0 units and a y component of -60.0 units.

(a) What are the two possibilities for its x component?

(b) Assuming the x component is known to be positive, specify the vector which, if you add it to the original one would give a resultant vector that is 70.0 units long and points entirely in the -x direction. (looking for Magnitude + Direction)
 
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Suppose that a vector has the components x and y, and a magnitude r. What's the equation that links x, y and r?
 
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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