What is the value of its x-component?

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To find the x-component of a velocity vector that is 55 degrees below the positive x-axis with a y-component of -13 m/s, a diagram can be helpful. By using trigonometric functions, specifically cosine for the x-component, the calculation can be performed. The relationship between the components can be expressed as: x-component = y-component / tan(angle). This approach will yield the correct x-component value. Understanding the geometry of the vector is crucial for solving the problem accurately.
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A velocity vector 55 degrees below the positive x-axis has a y-component of -13 m/s.

What is the value of its x-component?

Ive tried many different ways for this problem but i keep getting it wrong! help!
 
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bpatel4116 said:
A velocity vector 55 degrees below the positive x-axis has a y-component of -13 m/s.

What is the value of its x-component?

Ive tried many different ways for this problem but i keep getting it wrong! help!

Welcome to PF.

If your vector is pointing 55° down then draw the diagram.

You have a triangle - you have the angle - they tell you a side - then figure out the length of the hippopotamus.
 
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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