I dont understand how to get velocity from (6.00i - 1.00j) ?

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To find the velocity from the vector (6.00i - 1.00j) m/s, one must calculate its magnitude using the Pythagorean theorem. The magnitude is determined by taking the square root of the sum of the squares of its components, specifically √(6² + (-1)²). This results in a magnitude of approximately 6.08 m/s. Additionally, understanding vector mechanics is crucial for interpreting the direction and resultant of the vector. Familiarity with vector operations, including the dot product, is beneficial for solving such problems.
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I don't understand how to get velocity from (6.00i - 1.00j) ??

Homework Statement



Hey all, I have a problem where I am asked to find the velocity of something, but some data I am given says: "A 3.00kg object has a velocity (6.00i - 1.00j) m/s "

What is the velocity of that? I don't quite understand how to get vel from that vector.

Homework Equations





The Attempt at a Solution

 
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You are given the two components of the vector v. Phytagoras will tell you how to find the magnitude of the velocity vector.
 


I was reading over, and it says v^2 = v * v

?

so, is it v = square root 6?
 


Wot, no notes or textbook?
Whats 6squared + 1squared?

Look up how to find the magnitude and direction of a vector, given it's xy components. Or even how to find the resultant of two vectors at right angles.
 


I think that you were reading that the dot product (or scalar product) of the two vectors \underline{v} gives the MAGNITUDE of the vector \underline{v}.

i.e.\underline{v}.\underline{v}=v^{2}
 


Have you studied vector mechanics?
 
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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