Vector A has y-component Ay= +12.0A makes an angle of 32.0

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Vector A has a y-component of Ay = +12.0 A and makes a 32.0-degree angle counterclockwise from the positive y-axis. To find the x-component of A and its magnitude, trigonometric functions can be applied after drawing a diagram of the vector. The positive y-axis is defined as the upward direction, which helps in visualizing the vector's orientation. Understanding the orientation of the axes is crucial for accurately decomposing the vector into its components. Drawing the problem and applying trigonometry will simplify the calculations needed to solve for the x-component and magnitude.
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Vector A has y-component Ay= +12.0 . A makes an angle of 32.0 degrees counterclockwise from the + y-axis.

What is the x - component of A?

What is the magnitude of A?


How can I do this?
 
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You draw it, and then use trigonometry. (Always draw the problem!) Because you have learned about decomposition about vectors into components, right? This is more or less the same problem.
 


I don't understand the + y - axis
 


oldspice1212 said:
I don't understand the + y - axis

That would be the positive y axis, as in, going upwards, should you choose to define upwards as being positive.

Then, like Hypersphere said, you always need to draw a diagram, and take the proper trigonometric steps to figure out the components and magnitude.
 


Can you locate the y-axis and then tell what part of it is positive? It should be pretty easy to tell. And from there you can figure out where to draw the vector.
 
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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